Is this function differentiable at $(0,0)$? Let $f: \mathbb R^2  \to \mathbb R$ be the function
$$f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$$
where $(x,y) \neq  (0,0)$ and $f(0,0)=0$.
Is $f$ differentiable at $(0,0)$ and if so, how can I prove it?
 A: We will prove that $f$ is differentiable at $(0,0)$ with derivative $df_{(0,0)}=0$.
Using $|\sin z|\leq |z|$, we have 
$$
\frac{|f(x,y)|}{\sqrt{x^2+y^2}}\leq \frac{|x|^3|x+y|}{(x^2+y^2)^{3/2}}+\frac{|y|}{(x^2+y^2)^{1/2}}\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}.
$$
Now using Cauchy-Schwarz, we find
$$
\frac{|x|^3|x+y|}{(x^2+y^2)^{3/2}}\leq \frac{|x|^3\sqrt{2}\sqrt{x^2+y^2}}{(x^2+y^2)^{3/2}}=\frac{|x|^3\sqrt{2}}{x^2+y^2}\leq \frac{\sqrt{2}|x| (x^2+y^2)}{x^2+y^2}=\sqrt{2}|x|\leq \sqrt{2}\sqrt{x^2+y^2}
$$
So this term converges to $0$ as $\sqrt{x^2+y^2}$ tends to $0$.
Next observe that 
$$
\frac{|y|}{(x^2+y^2)^{1/2}}\leq \frac{\sqrt{x^2+y^2}}{(x^2+y^2)^{1/2}}=1
$$
and
$$
\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}=\frac{y^2}{x^2+y^2}\cdot|y\ln(x^2+y^2)|\leq \sqrt{x^2+y^2}\ln(x^2+y^2).
$$
Finally, 
$$
\frac{|y|}{(x^2+y^2)^{1/2}}\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}\leq \sqrt{x^2+y^2}\ln(x^2+y^2)
$$
so this term also converges to $0$ as $\sqrt{x^2+y^2}$ tends to $0$, using $\lim_{u\rightarrow 0^+}\sqrt{u}\ln u=0$.
So 
$$
\lim_{(x,y)\rightarrow(0,0)} \frac{|f(x,y)-f(0,0)-0|}{\sqrt{x^2+y^2}}=0
$$
which proves exactly our claim.
