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$$f(x,y)=\begin{cases} \frac{\ln{(1+x^2 y^2})}{x} & , & x \neq0 \\ 0 & , & x=0 \end{cases}$$

My idea is to prove that all partial derivatives are continuous on $\mathbb{R}^2$.

$$\frac{\partial f}{\partial x}(x,y)=\frac{2x^2y^2-\ln{(1+x^2y^2)(1+x^2y^2)}}{(1+x^2y^2)x^2} \qquad , \quad x\neq 0$$

$$\frac{\partial f}{\partial y}(x,y)=\frac1{x}\frac1{1+x^2y^2}2yx^2 \qquad , \quad x \neq 0$$

Now I want to find partial derivatives for $x=0$

$$\lim_{h \to 0} \frac{f(h,y)}{h}=\lim_{h \to 0} \frac{\ln{(1+h^2y^2)}}{h^2}$$

I don't know how to compute this limit. Am I doing something wrong? After that I would prove that partial derivatives are continuous at $x=0$.

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    Thank you, I totally forgot about it, but is the concept good? – 2017-02-12

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For the final limit (assuming earlier calculations are correct) you can use de L'Hopital $$ \frac{\partial f}{\partial x}(0,y)=\lim_{h \to 0} \frac{\ln{(1+h^2y^2)}}{h^2}= \lim_{h \to 0} \frac{2hy^2}{2h(1+h^2y^2)}= \lim_{h \to 0} \frac{y^2}{1+h^2y^2}=y^2 $$ On the other hand: $$ \frac{\partial f}{\partial x}(x,y)=\frac{2x^2y^2-\ln{(1+x^2y^2)(1+x^2y^2)}}{(1+x^2y^2)x^2} = \frac{2y^2}{1+x^2y^2} - \frac{\ln{(1+x^2y^2)}}{x^2} $$ so $$ \lim_{x\rightarrow 0}\frac{\partial f}{\partial x}(x,y) = 2y^2-y^2=y^2=\frac{\partial f}{\partial x}(0,y) $$

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    Thank you, do I have to check partial derivative for $y$ at $0$? I think yes – 2017-02-12
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    Yes you need to do that at $(0,y)$ but it will be a much easier calculation. – 2017-02-12
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    I get $$\lim_{h \to 0} \frac{f(0,y+h)}{h}=\frac{0}{0}$$ – 2017-02-12
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    @RushThaMan You will need to use L'Hopital then – 2017-02-12