How to prove $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$

If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $\tau(n)=\sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.

Recall that if $f$ is a multiplicative function, and $$g(n)=\sum_{d|n}f\left(\frac{n}{d}\right),$$ then $g$ is a multiplicative function.

From this you can deduce that both sides of your identity are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.

For $n=p^k$, the left-hand side is $1^3+2^3+\cdots +k^3+(k+1)^3$ (since $p^i$ has $i+1$ positive divisors for each $i \geq 0$). The right-hand side is $(1+2+\cdots+k+(k+1))^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.