Let $f$ be any differentiable function. Show that $w(x,y) = f(x y^2)$ satisfies $2x \frac{\partial w}{\partial x} - y \frac{\partial w}{\partial y} = 0.$

Solution

To show $w$ satisfies that equation, we compute the left hand side and verify that it is 0.

The derivative of $w$ with respect to $x$ can be computed by the single-variable chain rule: $$\frac{\partial w}{\partial x} = f'(xy^2) y^2.$$

The derivative with respect to $y$ can also be computed by the single-variable chain rule: $$\frac{\partial w}{\partial x} = f'(xy^2) 2xy.$$

Plugging into the equation, we see that $$ 2x \frac{\partial w}{\partial x} - y \frac{\partial w}{\partial y} = f'(x y^2) 2xy^2 - f'(x y^2) 2xy^2 = 0.$$