Differentiability of $x^2\log(x^4+y^2)$ at $(0,0)$ Based on this question, I have the function
$$
f(x,y)=\begin{cases}
x^2\log(x^4+y^2), & (x,y)\in\mathbb{R}^2\setminus\{(0,0)\},\\
0,                & (x,y)=(0,0).
\end{cases}
$$
I would like to study its continuity and differentiability at $(0,0).$
Continuity
For the continuity, I see that I can rewrite the expression $x^2\log(x^4+y^2)$ as
$$
\sqrt\frac{x^4}{x^4+y^2}\cdot\sqrt{x^4+y^2}\log(x^4+y^2),
$$
and given that:


*

*$\sqrt\frac{x^4}{x^4+y^2}$ is bounded:
$$
0\leq\sqrt\frac{x^4}{x^4+y^2}\leq1;
$$

*for $\sqrt{x^4+y^2}\log(x^4+y^2)$ I can use the known limit
$$
\lim_{t\to0}t^\alpha\log t=0,\qquad\forall\alpha>0;
$$


I conclude that 
$$
\lim_{(x,y)\to(0,0)}f(x,y)=0,
$$
so the function is continuous in $(0,0).$
Existence and continuity of derivative with respect to $x$
Something similar can be done for the derivative with respect to $x$ in $(0,0),$ in fact
$$
f'_x(0,0)=\lim_{(x,y)\to(0,0)}\frac{x^2\log(x^4+y^2)-0}{x-0}=\lim_{(x,y)\to(0,0)}x\log(x^4+y^2)
$$
and the expression $x\log(x^4+y^2)$ could be written as
$$
\operatorname{sign}x\cdot\sqrt[4]\frac{x^4}{x^4+y^2}\cdot\sqrt[4]{x^4+y^2}\log(x^4+y^2),
$$
whose limit, as before, is $0.$
As for the limit of the derivative function with respect to $x$, it is
$$
\lim_{(x,y)\to(0,0)}f'_x(x,y)=\lim_{(x,y)\to(0,0)}\left(2x\log(x^4+y^2)+\frac{4x^5}{x^4+y^2}\right)
$$
and the first term of the sum is as before, while the second can be written as the product of $4x$, that goes to $0$, by a bounded ratio, so the limit is $0$ and I can conclude that $f'_x(x,y)$ is continuous at $(0,0).$  
Existence and continuity of derivative with respect to $y$
For the derivative with respect to $y$ the things are different, I have
$$
f'_y(0,0)=\lim_{(x,y)\to(0,0)}\frac{x^2\log(x^4+y^2)-0}{y-0}=\lim_{(x,y)\to(0,0)}\frac{x^2}{y}\log(x^4+y^2)
$$
that I am not able to rewrite in a form that is simpler to manage. Moreover, if I make the limit along the curves $y=x^2$ and $y=-x^2$ I get
$$
\lim_{x\to0}\frac{x^2}{x^2}\log(2x^4)=-\infty\\
\lim_{x\to0}\frac{x^2}{-x^2}\log(2x^4)=+\infty
$$
so the derivative with respect to $y$ does not exist in $(0,0),$ and the function cannot be differentiable in $(0,0).$
The question
Given all that, my question is:
this question is wrong saying $f$ is differentiable, or I am making some mistake?
Also the graphics of $f$ seems rather smooth around $(0,0)$:

 A: My error was to use the wrong formula
$$
f'_x(0,0)=\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)}{x-0}
$$
instead of the correct one
$$
f'_x(0,0)=\lim_{x\to0}\frac{f(x,0)-f(0,0)}{x-0}
$$
and the same error for $f'_y(0,0)$.
With this correction I get
$$
f'_x(0,0)=\lim_{x\to0}x\log x^4=0
$$
and 
$$
f'_y(0,0)=\lim_{y\to0}0=0
$$
Now
$$
\lim_{(x,y)\to(0,0)}f'_y(x,y)=\lim_{(x,y)\to(0,0)}\frac{2x^2y}{x^4+y^2}
$$
and this is $\pm1$ along the lines $y=\pm x^2$, so I cannot say if $f$ is differentiable at $(0,0)$ by continuity of partial derivatives, I should use the definition, i.e.
$$
\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{\sqrt{x^2+y^2}}=0
$$
so I should calculate
$$
\lim_{(x,y)\to(0,0)}\frac{x^2\log(x^4+y^2)}{\sqrt{x^2+y^2}}=
\lim_{(x,y)\to(0,0)}\sqrt\frac{x^2}{x^2+y^2}\cdot\sqrt[4]\frac{x^4}{x^4+y^2}\cdot\sqrt[4]{x^4+y^2}\log(x^4+y^2)=0,
$$
because the product of two bounded functions and a third one that goes to $0$.
Finally I conclude that $f$ is differentiable at $(0,0)$.
