$\left\{\begin{matrix} x^2 x'=\sin^2(x^3-3t) \\ x(0)=1 \end{matrix}\right.$ 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, $x=\sqrt[3]{3t+\tan^{-1}(tan(1) -3t) }$
