Why can't I use the chain rule on $f(x,y) = \frac{xy^2}{x^2 + y^2}$?

Question

The function is $f(x,y) = \frac{xy^2}{x^2 + y^2}$ and $0$ if $(0,0)$. So $f(x,y)$ is continous, the partial derivatives exist, and the partial derivatives are also continuous (the limit as $(x,y) \rightarrow (0,0)$ is $0$). Thus $f(x,y)$ is differentiable, but why can't the chain rule be applied to $f \circ \textbf{g}$ where $\textbf{g}(t) = (at, bt)$? The textbook says you can't apply the chain rule here because $f$ is not differentiable, but obviously it is so I am confused. Then it says something like $(f \circ \textbf{g})'(0) = \frac{ab^2}{a^2 + b^2}$ and $\nabla f(0,0) \cdot \textbf{g}'(0) = 0$, but I don't know what the gradient has anything to do with the chain rule.

Answer 1

The partial derivatives are not continuous at $(0, 0)$. You have $f_x(0, 0) = 0$ and

$$ f_x(x, y) = \frac{y^2}{x^2 + y^2} - \frac{2 x^2 y^2}{(x^2 + y^2)^2} \text{ for } (x,y) \ne (0, 0) $$

so that $$ \lim_{y \to 0} f_x(0, y) = 1 \ne 0 = f_x(0, 0) \quad . $$

Therefore you cannot conclude that $f$ is differentiable at $(0, 0)$.