Finding solution of non linear DE $x\ddot{x}-\dot{x}^2=1$

Yes, there is a way. Dividing both sides of the differential equation by $x^2$ gives

$$\frac{x\ddot{x} - \dot{x}^2}{x^2} = \frac{1}{x^2}$$

or

$$\frac{d}{dt}\left(\frac{\dot{x}}{x}\right) = \frac{1}{x^2}.$$

Since $\displaystyle \frac{\dot{x}}{x} = \frac{\mathrm{d}}{\mathrm{d}t}(\log x)$, we have

$$\frac{d^2}{dt^2}(\log x) = \frac{1}{x^2}.$$

Now set $x = e^u$, so that

$$\ddot{u} = e^{-2u}.$$

Multiplying both sides by $\dot{u}$,

$$\ddot{u}\dot{u} = e^{-2u}\dot{u}$$

or

$$\frac{d}{dt}\left(\frac{\dot{u}^2}{2}\right) = -\frac{d}{dt}\frac{e^{-2u}}{2}.$$

By integration,

$$ \dot{u}^2 = C - e^{-2u}$$

where $C$ is a constant. Hence $\dot{u} = \pm \sqrt{C - e^{-2u}}$.

By separation of variables,

$$\int \frac{du}{\sqrt{C - e^{-2u}}} = \pm\int dt.$$

Now

$$\int \frac{du}{\sqrt{C - e^{-2u}}} = \int \frac{e^u\, du}{\sqrt{Ce^{2u} - 1}}.$$

Setting $\sqrt{C}e^u = \cosh \theta$, we get

$$\int \frac{e^u\, du}{\sqrt{Ce^{2u} -1}} = \int \frac{\sinh\theta \,d\theta}{\sqrt{C}\sinh \theta} = \frac{1}{\sqrt{C}}(\theta + D) = \frac{1}{\sqrt{C}}\cosh^{-1}(\sqrt{C}e^u) + \frac{D}{\sqrt{C}}.$$

It follows that

$$\frac{1}{\sqrt{C}}\cosh^{-1}(\sqrt{C}e^u) = \pm t + A,$$

where $A$ is a constant. Since $e^u = x$,

$$\frac{1}{\sqrt{C}}\cosh^{-1}(\sqrt{C}x) = \pm t + A$$

or

$$x(t) = \frac1{\sqrt{C}}\cosh(\sqrt{C}(t + A))$$

which we can write in the form

$$x(t) = c_1 \cosh\left(\frac{t - c_2}{c_1}\right) .$$

Let $\dot x=p$. Then $$ \ddot x=\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=p\,\frac{dp}{dx}. $$ The equation becomes $$ x\,p\,\frac{dp}{dx}-p^2=1\implies\frac{p\,dp}{p^2+1}=\frac{dx}{x}. $$ Integration results in $$ p^2+1=C\,x^2\implies {\dot x\,}^2=c_1\,x^2-1\implies \frac{dx}{\sqrt{c_1\,x^2-1}}=\frac{dt}{t}. $$ The solution is $$ \int\frac{dx}{\sqrt{c_1\,x^2-1}}=\log t+c_2. $$