# Total derivative of function with not continuous derivatives

I was wondering if the following function is totally differentiable in $$(0, 0)$$: $$f(x, y) = \frac{x^2y}{x^2+y^2}$$ and $$f(0,0) = 0$$. I know that the partial derivatives are $$f_x(x, y)= \frac{2xy^3}{(x^2+y^2)y^2}, \quad f_y(x, y) = \frac{x^4-x^2y^2}{(x^2+y^2)^2}.$$ Is it correct that both of these partial derivatives are not continuous in $$(0,0)$$? For $$f_x(x,y)$$, one can consider the limits $$\lim_{h\to 0}f_x(h, h)=1/2$$ and $$\lim_{h\to 0}f_x(h, 0) = 0$$, so this derivative can't be continuous, similarly for $$f_y(x,y)$$, which also can't be continuous, and in summary, $$f$$ can't be totally differentiable in $$(0,0)$$.

All I'm wondering is if my thoughts are correct here. Thanks!

We have $$f(x,0) = 0$$ and $$f(0,0) = 0$$. Then, $$f'_x(0,0) = \lim_{x\to 0}\frac{f(x,0) - f(0,0)}{x-0} = 0$$ Similarly, we get $$f'_y(0,0) = \lim_{y\to 0}\frac{f(0,y) - f(0,0)}{y-0} = 0$$ Hence we need to check if the following equality is true: $$\lim_{x\to0 \\ y\to0} \frac{f(x,y) - f(0,0) - 0 - 0}{\sqrt{x^2 + y^2}} \overset ? = 0$$ which is equivalent to $$\lim_{x\to0 \\ y\to0} \frac{\frac{x^2y}{x^2+y^2}}{\sqrt{x^2+y^2}}= \lim_{x\to0 \\ y\to0} \frac{x^2y}{(x^2+y^2)^{3/2}} \overset ?= 0$$ Can you take it from here?
• Thanks! I understand the idea, but why do we need to take $\sqrt{x^2+y^2}$ in the third step? Everything else is clear Commented Jan 16, 2021 at 12:55
• @Gecko1111 The definition of differentiability says: $$\Delta f (\textbf{x}) - \nabla f (\textbf{x})\Delta \textbf x = o(\rho (\textbf{x} + \Delta\textbf{x}, \textbf x))$$ where $\textbf{x} = (x,y)$ and $\rho(\textbf{x} + \Delta\textbf{x}, \textbf x) = \sqrt{\Delta x^2+ \Delta y^2}$. Commented Jan 16, 2021 at 13:08
• Oh right, I did not see that the root is the norm of $(\Delta x ,\Delta y)$. I calculated the limit using $y =ax, a \in \mathbb{R}$, then the limit becomes $\lim_{x \to 0} \frac{a}{(1+a^2)^{3/2}}$. So that $f$ is not totally differentiable. Commented Jan 16, 2021 at 17:45