From system of coupled ODEs to separable ODE

Notice that:

$$\dot{x} = y \Rightarrow dx = ydt,$$

and

$$\dot{y} = -x^3 \Rightarrow dy = -x^3 dt.$$

Therefore:

$$\frac{dy}{dx} = \frac{-x^3dt}{ydt} = -\frac{x^3}{y}$$.

Hint: $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$$

We know that $\dot{y}=-x^3$. Divide this by $\dot{x}=y$ and we get $\frac{\dot{y}}{\dot{x}}=\frac{-x^3}{y}$. But $\frac{\dot{y}}{\dot{x}}$ is just $\frac{dy}{dx}$ so we're done.