Answer. Solution does not exist.

Let us substitute $x = y = 0$ at the first condition $\Rightarrow f(f(0)) = 0$.
Then when $x=0$ and when $x=f(0)$ we obtain such equations:
$$f(y^2 + f(y)) = y f(y) + y \quad \text{ta} \quad f(f^3(0) + y^2 + f(y)) = y f(y) + y.$$
Since $f$ is strictly increasing:
$$f(y^2 + f(y)) = f(f^3(0) + y^2 + f(y)) \Rightarrow f(0) = 0.$$
If we substitute $x=0$ and $y=0$, we obtain
$$\begin{aligned} f(y^2 + f(y)) &= y f(y) + y \quad \text{ta} \quad f(x^3) = x^2 f(x) \\ f(x^3 + y^2 + f(y)) &= f(x^3) + f(y^2 + f(y)). \end{aligned}$$
Since $y^2 + f(y)$ can equal any positive value (from the second condition), the last equation can be rewritten in such a way: $f(x+t) = f(x) + f(t)$, where $x \in \mathbb{R}, t \ge 0$.

Therefore, $f$ is increasing and satisfies Cauchy's functional equation, thus $f(x) = kx$ for some real $k$. After checking the first condition it is clear that only case $k=1$ is correct, but such a function does not satisfy the second condition.