Is $\eta^{24}(\tau)\,j(\tau) = {E_4}^3(q)$?

Yes, these three relations are all true. Since $\eta^{24}$ is the weight 12 level 1 cusp form $\Delta$, you can write them as relations between level 1 modular forms, and these are easy to check because the relevant modular form spaces have small finite dimensions.

The first and third identity are easy to prove. Recall that the modular discriminant $\Delta = \Delta(\omega_1, \omega_2)$ is defined by $$\Delta = g_2^3 - 27g_3^2,$$ where $g_2 = g_2(\omega_1, \omega_2) = 60G_4(\omega_1, \omega_2)$ and $g_3 = g_3(\omega_1, \omega_2) = 140G_6(\omega_1, \omega_2)$ are the Weierstrassian invariants. Moreover, we define the Klein invariant $J = J(\omega_1, \omega_2)$ and the $j$-function by $$J = \frac{g_2^3}{\Delta} \quad \text{and} \quad j = 12^3 J,$$ respectively.

Let $\lambda \ne 0$. Because $g_2$ and $g_3$ are of order $-4$ and $-6$, respectively, we have $$g_2(\lambda \omega_1, \lambda \omega_2) = \lambda^{-4} g_2 \quad \text{and} \quad g_3(\lambda \omega_1, \lambda \omega_2) = \lambda^{-6} g_3,$$ and $$\Delta(\lambda \omega_1, \lambda \omega_2) = \lambda^{-12} \Delta,$$ accordingly. Moreover, since $g_2^3$ and $\Delta$ are of the same order, $J(\lambda \omega_1, \lambda \omega_2) = J(\omega_1, \omega_2)$. In particular, $$g_2(\tau) = \omega_1^4 g_2, \quad g_3(\tau) = \omega_1^6 g_3, \quad \text{and} \quad \Delta(\tau) = \omega_1^{12} \Delta$$ On the other hand, $$J = J(\tau) \quad \text{and} \quad j = j(\tau),$$ that is to say, both $J$ and $j$ are functions of $\tau$ alone.

Using the classical identity $\Delta(\tau) = (2\pi)^{12} \eta^{24}(\tau)$, where $\eta(\tau)$ is the Dedekind eta function defined by $$\eta(\tau) = q^{1/12} \prod_{n = 1}^\infty (1 - q^{2n}), \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ we obtain $$\eta^{24}(\tau) = \frac{\Delta(\tau)}{(2\pi)^{12}} = \frac{\omega_1^{12} g_2^3}{(2\pi)^{12} J(\tau)} = \frac{12^3 \omega_1^{12} g_2^3}{(2\pi)^{12} j(\tau)}.$$ In view of a known result, $$E_{2k}(q) = 1 - \frac{4k}{B_{2k}} \sum_{n = 1}^\infty \frac{n^{2k - 1} q^{2n}}{1 - q^{2n}}, \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ where $B_{2k}$ are the Bernoulli numbers and $E_{2k}(q) = G_{2k}(\tau)/2\zeta(2k)$, we get $$g_2(\tau) = 60G_4(\tau) = \frac{4\pi^4 E_4(q)}{3},$$ or, what is the same thing, $$g_2 = \omega_1^{-4} \frac{4\pi^4 E_4(q)}{3},$$ Therefore, $$\eta^{24}(\tau) = \frac{12^3 \omega_1^{12}}{(2\pi)^{12} j(\tau)} \left(\omega_1^{-4} \frac{4\pi^4 E_4(q)}{3}\right)^3 = \frac{E_4^3(q)}{j(\tau)}.$$ Lastly, using the fact that $E_8(q) = E_4^2(q)$ leads to $$\eta^{48}(\tau) = \frac{E_8^3(q)}{j^2(\tau)}.$$

Now, the second identity follows from the first and the fact that $$J(\tau) = \frac{E_4^3(q)}{E_4^3(q) - E_6^2(q)}.$$ Indeed, we have $$E_6^2(q) = E_4^3(q) - 12^3 \frac{E_4^3(q)}{j(\tau)},$$ but in view of the first identity this becomes $$E_6^2(q) = (j(\tau) - 12^3)\eta(\tau)^{24},$$ and the result follows.

These relations are easily proved if we use the nome $q = e^{2\pi i\tau}$ and use Ramanujan notation of $P(q), Q(q), R(q)$ also.

We have \begin{align} \eta(\tau) &= \eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n})\tag{1}\\ E_{2}(\tau) &= P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}}\tag{2}\\ E_{4}(\tau) &= Q(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}}\tag{3}\\ E_{6}(\tau) &= R(q) = 1 - 504\sum_{n = 1}^{\infty}\frac{n^{5}q^{n}}{1 - q^{n}}\tag{4}\\ E_{8}(\tau) &= Q^{2}(q) = 1 + 480\sum_{n = 1}^{\infty}\frac{n^{7}q^{n}}{1 - q^{n}}\tag{5}\\ j(\tau) &= \frac{1728E_{4}^{3}(\tau)}{E_{4}^{3}(\tau) - E_{6}^{2}(\tau)} = \frac{1728Q^{3}(q)}{Q^{3}(q) - R^{2}(q)}\tag{6} \end{align} Except for the fact $E_{8}(\tau) = Q^{2}(q) = E_{4}^{2}(\tau)$ mentioned in equation $(5)$ all the equations above are definitions of various functions of $q$ and corresponding functions of $\tau$.

It is easy to see that $$P(q) = 24q\frac{d}{dq}\{\log \eta(q)\}\tag{7}$$ And Ramanujan gives the fundamental differential equations \begin{align} q\frac{dP(q)}{dq} &= \frac{P^{2}(q) - Q(q)}{12}\tag{8}\\ q\frac{dQ(q)}{dq} &= \frac{P(q)Q(q) - R(q)}{3}\tag{9}\\ q\frac{dR(q)}{dq} &= \frac{P(q)R(q) - Q^{2}(q)}{2}\tag{10} \end{align} From these differential equations it is easy to see that \begin{align} q\frac{d}{dq}\{Q^{3}(q) - R^{2}(q)\} &= 3Q^{2}(q)q\frac{dQ(q)}{dq} - 2R(q)q\frac{dR(q)}{dq}\notag\\ &= 3Q^{2}(q)\frac{P(q)Q(q) - R(q)}{3} - 2R(q)\frac{P(q)R(q) - Q^{2}(q)}{2}\notag\\ &= P(q)Q^{3}(q) - R(q)Q^{2}(q) - P(q)R^{2}(q) + R(q)Q^{2}(q)\notag\\ &= P(q)\{Q^{3}(q) - R^{2}(q)\}\notag \end{align} and therefore $$q\frac{d}{dq}\{\log(Q^{3}(q) - R^{2}(q))\} = P(q)\tag{11}$$ Now comparing $(7)$ and $(11)$ we get $$Q^{3}(q) - R^{2}(q) = A\eta^{24}(q)$$ where $A$ is constant. Comparing coefficients of $q$ on both sides we can see that $$A = 3\cdot 240 + 2\cdot 504 = 1728$$ so that $$Q^{3}(q) - R^{2}(q) = 1728\eta^{24}(q)\tag{12}$$ Using this identity along with equation $(6)$ we get the first identity in question $$\eta^{24}(\tau) = \frac{E_{4}^{3}(\tau)}{j(\tau)}$$ and third identity in question is square of the first identity and the fact that $E_{8}(\tau) = E_{4}^{2}(\tau)$ as noted in equation $(5)$.

The second identity of question easily follows from equations $(6)$ and $(12)$.