$\left\{\begin{matrix} x^2 x'=\sin^2(x^3-3t) \\ x(0)=1 \end{matrix}\right.$

I have to solve the following Cauchy's problem:

$$ \left\{ \begin{align} & x^2 x'=\sin^2(x^3-3t) \\ & x(0)=1 \end{align} \right. $$

First of all I've tried to identify if it is homogeneous, linear, exact, euler's ... but I can't recognise it. So I don't know which procedure to follow to solve it...

I think that my problem is that inside the $\sin$ there are both $x$ and $t$.

Could anyone help me?

$\left\{\begin{matrix} x^2 x'=\sin^2(x^3-3t) \space \space ......(1)\\ x(0)=1 \end{matrix}\right.$

Let, $x^3-3t=y$

$3x^2\frac{dx}{dt}-3=\frac{dy}{dt}$

Now, (1) becomes,

$\frac{dy}{dt}=3( \sin^2(y)-1)$

$\implies \frac{dy}{dt}=-3(\cos^2(y))$

$\implies \frac{dy}{cos^2(y) }=-3dt$

$\implies \tan(y) =-3t+c$

Hence, $\tan(x^3-3t) =-3t+c$

Using initial condition $x(0) =1$

$c=\tan(1) $

Hence, $x=\sqrt[3]{3t+\tan^{-1}(tan(1) -3t) }$