Problem

The variables $x$ and $y$ are such that $y = 2$ when $x = -1$ and

$$x^2y + (x + y)^3 = 3.$$

(a) Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{4}$ when $x = -1$.

(b) Find the value of $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ when $x = -1$.