$$\frac{\mathrm{d}(y^3-4x)}{\mathrm{d}x}=\frac{\mathrm{d}2}{\mathrm{d}x}$$ Note that $\text{RHS}$ (Right Hand Side) is $0$, since $2$ is constant. By the linearity of the differential operator, we can write $$\frac{\mathrm{d}(y^3-4x)}{\mathrm{d}x}=\frac{\mathrm{d}y^3}{\mathrm{d}x} -4 \frac{\mathrm{d}x}{\mathrm{d}x}=\frac{\mathrm{d}2}{\mathrm{d}x}$$ By the chain rule, we can write $$\frac{\mathrm{d}y^3}{\mathrm{d}y} \times \frac{\mathrm{d}y}{\mathrm{d}x} -4 \frac{\mathrm{d}x}{\mathrm{d}x}=0$$ So $$3y^2 \frac{\mathrm{d}y}{\mathrm{d}x}-4=0$$ So we have $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4}{3y^2} \quad \left( \text{given that} y \neq 0 \right)$$
WRT $x$.
