Let $f(0) = \lambda$. Put $x = y = 0$ in (1) to get $f(\lambda) = 0$. Taking $y = \lambda$, we get $f(x^3) = x^2 f(x) + \lambda$. Put $x = \lambda, y = 0$ in (1) and we get
$$f(\lambda^3 + \lambda) = \lambda^2 f(\lambda). \qquad (2)$$
But $x = \lambda, y = \lambda$ gives $f(\lambda^3) = \lambda$. Now put $x = \lambda, y = \lambda^3$ and we get
$$f(\lambda^3 + \lambda) = \lambda^2 f(\lambda) + \lambda^3. \qquad (3)$$
Comparing (2) and (3), we get $\lambda = 0$. Therefore
$$f(x^3) = x^2 f(x), \quad f(f(x)) = x,$$
for all $x$. Now changing $y$ to $f(y)$, we get
$$f(x^3 + y) = x^2 f(x) + f(y) = f(x^3) + f(y)$$
for all $x, y$. Since $x \to x^3$ is a bijection of $\mathbf{R}$ onto $\mathbf{R}$, we obtain
$$f(x + y) = f(x) + f(y), \quad f(f(x)) = x$$
for all $x, y$.
Consider $f(x^3) = x^2f(x)$. Replacing $x$ by $x+1$, we obtain
$$f(x^3 + 3x^2 + 3x + 1) = (x^2 + 2x + 1)f(x + 1).$$
Using the additivity of $f$, we get
$$\begin{aligned} f(x^3) + 3f(x^2) + 3f(x) + f(1) &= (x^2 + 2x + 1)(f(x) + f(1)) \\ &= x^2 f(x) + 2x f(x) + f(x) + x^2 f(1) + 2x f(1) + f(1). \end{aligned}$$
This reduces to
$$3f(x^2) + 2f(x) = cx^2 + 2cx + 2xf(x) \quad (4)$$
for all $x \in \mathbb{R}$. Replace $x$ by $(x+1)$ and expand using additivity, we get
$$3f(x^2) + 6f(x) = cx^2 + 6cx + 2xf(x) \quad (5).$$
Comparing (4) and (5), we see that $f(x) = cx$. Replacing $x$ by $f(x)$, we get $x = cf(x)$. Together we get $c^2x = x$. Hence $c^2 = \pm 1$. Therefore $f(x) = \pm x$.