Show that $g(x,y) = x^2y^2\log(x^2+y^2), 0$ is differentiable in (0,0)

I have the following function: $$g(x,y) = \left\{\begin{array}{ll} x^2y^2\log(x^2+y^2), & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{array}\right.$$

I've already calculated the partial derivatives and showed that $g$ is partially differentiable: $$\lim_{h\rightarrow0} \frac{g(0+h,0) - g(0,0)}{h} = \frac{0-0}{h} \\ \lim_{h\rightarrow0} \frac{g(0,0+h) - g(0,0)}{h} = \frac{0-0}{h} = 0$$

Now I want to show that $g$ is differentiable in $(x,y)=(0,0)$. This is my try so far: \begin{align} &\lim_{h\rightarrow0} \frac{g(0+h) - g(0,0) - g'(0,0)h}{|h|} \\ &= \lim_{h\rightarrow0} \frac{h_1^2h_2^2 \log(h_1^2+h_2^2)}{\sqrt{h_1^2+h_2^2}} \\ \end{align}

But now I don't know how to continue and show that the limit is $0$.

Since the partial derivatives are both $0$, the derivative, if it exists, is $0$. Now it's a “squeezing question”: $$\frac{x^2y^2\log(x^2+y^2)}{\sqrt{x^2+y^2}}= \frac{x^2y^2}{(x^2+y^2)^2} (x^2+y^2)^{1/2} \bigl((x^2+y^2)\log(x^2+y^2)\bigr)$$ Prove that $$-1\le\frac{ab}{(a+b)^2}\le 1$$ for $a,b>0$ and recall that $\lim_{t\to0^+}t\log t=0$.
Transforming $h=(h_1,h_2)$ space into polar coordinates ($h_1=r\cos\theta,h_2=r\sin\theta$): $$\lim_{h\to0}\frac{h_1^2h_2^2\ln(h_1^2+h_2^2)}{\sqrt{h_1^2+h_2^2}}=\lim_{h\to0}\frac{(r\cos\theta)^2(r\sin\theta)^2\ln r^2}{\sqrt{r^2}}=\lim_{h\to0}2r^3\cos^2\theta\sin^2\theta\ln r$$ But then $$\lim_{h\to0}|2r^3\cos^2\theta\sin^2\theta\ln r|\le\lim_{r\to0}|2r^3\ln r|=0$$ so the original limit is also 0 from all directions.