Show $f(x,y)=x^2\log(x^4+y^2)$ is differentiable at $\vec 0$ I have to show that $f$ is differentiable at $\vec 0$, where
$$
f(x,y)=x^2\log(x^4+y^2),
$$
and $f(0,0)=0$.
I’ve already shown that $f$ is continuous at $\vec 0$. I started off by calculation the first partial derivative:
$$
D_1f(\vec 0)=\lim_{t\to 0}t\log t^4.
$$
However, I don’t know how to calculate this limit even. I looked at the plot, and it seems that $D_f(\vec 0)=D_2f(\vec 0)=0$, so apparently $t$ goes faster to zero then $\log t^4$ goes to minus infinity. How can I show this? Can I use Taylor? Should I evaluate then at $x=1$? This would yield:
$$
\log x=(x-1)-\frac{(x-1)^2}{2}+O((x-1)^3).
$$
Is this the way to go? I've never expanded $\log x$ before like this, and I'm unsure if it's correct.

$t\log t^4=4t\log t\to 0$ as $t\to 0_+$  is a standard limit from high school.
