# Partial derivative - problem

I have function defined as $f(x,y)=\sqrt[3]{x^2+y}\cdot \ln{(x^2+y^2)}, f(0,0)=0$. I am evaluating derivatives:

$\frac{\partial f}{\partial x}(x,y)=\frac{1}{3}\cdot\frac{2x\ln{(x^2+y^2)}}{\sqrt[3]{(x^2+y)^2}}+\frac{2x\sqrt[3]{x^2+y}}{x^2+y^2}$ and $\frac{\partial f}{\partial y}(x,y)=\frac{1}{3}\frac{\ln{(x^2+y^2)}}{\sqrt[3]{(x^2+y)^2}} + \frac{2y\sqrt[3]{x^2+y}}{x^2+y^2}$.

I hope I haven't made any mistakes, however what troubles me is $[0,0]$, in this point function is defined as $f(0,0)=0$. How do I treat it(Do I have to differentiate it as 'different' function)?

Lastly, I am trying to determine what happens when $y=-x^2$, I tried to find out by using limits:

Let $y_o=-x_0^2$, then we have $\frac{\partial f}{\partial x}(x_0,y_0)=\lim_{x \rightarrow x_0}2x(\frac{\ln{(x^2+y_0^2)}}{3\sqrt[3]{(x^2+y_0)^2}}+\frac{\sqrt[3]{x^2+y_0}}{x^2+y_0^2})=\lim_{x \rightarrow x_0}2x(\frac{\ln{(x^2+x_0^4)}}{3\sqrt[3]{(x^2-x_0^2)^2}}+\frac{\sqrt[3]{x^2-x_0^2}}{x^2+x_0^4})$

I can't get rid of $x^2-x_0^2$ in denominator, can you please help me with this problem? I am solving this kind of problems on my own for the first time, and also please don't use Taylor series, which I don't know. Thank you.

To compute partial derivatives at a ceratin point we can use the definition of partial derivatives.For example, $$f_x(0,0)=\lim_{h\rightarrow 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\rightarrow 0}\frac{h^{2/3}\ln (h^2)}{h}=\lim_{h\rightarrow 0}\frac{\ln h^2}{h^{1/3}}.$$ This limit does not exists as $h\rightarrow 0$. (It approaches infinity or minus infinity). So at $(0,0)$ partial derivative $f_x$ is not defined at (0,0). You can check $f_y(0,0)$ in a similar way.