Prove a multivariable function is differentiable 
Prove $f(x,y)=x^2+2x+y^2$ is differentiable in $\ \mathbb{R}^2$.

I need to show that there exists a linear transformation such that $\lim_{h\rightarrow0}\frac{(x+h)^2+2(x+h)+(y+h)^2-L(h)}{||h||}=0.$
$\frac{(x+h)^2+2(x+h)+(y+h)^2-(2x+2+2y)}{||h||}=\frac{x^2+2xh +h^2+2x + 2h+y^2+2yh+h^2-2x-2-2y}{||h||}=?$
$\frac{\partial{f(x,y)}}{\partial{x}}$=2x+2 and $\frac{\partial{f(x,y)}}{\partial{y}}$=2y
 A: $\newcommand{\dd}{\partial}$Hint: The partials of $f$ are the coefficients of the linear function $L_{(x, y)} = L$. That is,
$$
L(h, k) = \frac{\dd f}{\dd x}(x, y)\, h + \frac{\dd f}{\dd y}(x, y)\, k.
$$
The "variables" in $L$ are $(h, k)$, the displacements from $(x, y)$. The goal is to show
$$
\lim_{(h, k) \to (0, 0)} \frac{|f(x + h, y + k) - f(x, y) - L(h, k)|}{\sqrt{h^{2} + k^{2}}} = 0.
$$
A: I appreciate your help, Andrew D. Hwang. I've been trying to apply the same method to function $f(x,y)$
=\begin{cases} 
                 x^2sin(\frac{1}{x})+y^2, & x\neq0 \\
     \
                 y^2, & x=0
                \end{cases} to prove it's differentiable at (0,0). This time
$\frac{\partial{f(0,0)}}{\partial{x}}=0=\frac{\partial{f(0,0)}}{\partial{y}}$
and 
$$
 \frac{|f(0 + h, 0 + k) - f(0, 0)|}{\sqrt{h^{2} + k^{2}}} =\frac{|f(h,k) - f(0, 0)|}{\sqrt{h^{2} + k^{2}}}=\frac{|h^2sin(1/h)+k^2 - 0^2|}{\sqrt{h^{2} + k^{2}}}=\frac{|h^2sin(1/h)+k^2 |}{\sqrt{h^{2} + k^{2}}}  
$$aand I am stuck again. I have tried to canlcel out the denominator.
