If and , then prove that .

Given $x = 3z + 1$ and $y = f(x)$. Differentiating $x$ with respect to $z$ gives $\frac{dx}{dz} = 3$. By the chain rule: $\frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = 3\frac{dy}{dx}$. Differentiating again with respect to $z$: $\frac{d^2y}{dz^2} = \frac{d}{dz}\left(3\frac{dy}{dx}\right) = 3\frac{d}{dz}\left(\frac{dy}{dx}\right) = 3\left[\frac{d}{dx}\left(\frac{dy}{dx}\right) \cdot \frac{dx}{dz}\right] = 3\left(\frac{d^2y}{dx^2}\right)(3) = 9\frac{d^2y}{dx^2}$. Hence, $9\frac{d^2y}{dx^2} = \frac{d^2y}{dz^2}$.