Partial derivatives with transformation matrix I have the following function:
$ f(x,y) = 2xy + \frac{x}{y}$
I need do determine if it is differentiable at $ a = (1,1) $.
Calculating the partial derivatives and plugging the values, I get that the linear transformation corresponds with multiplying by :
$ [ 2y + \frac{1}{y}, 2x - \frac{x}{y^2}] = [ 3, 1]  $
so as $[h,k] \to [0,0]$, I have:  
$f(1 + h, 1 +k) - f(1,1) - \lambda([h,k]) = \frac{k^2+hk+2hk^2}{1+k} $
thus I need to show:
$\lim \limits_{[h,k] \to [0,0]} \frac{\frac{k^2+hk+2hk^2}{1+k}}{||[h,k]||} = 0 $
when $||[h,k]||$ is the euclidean norm.
This is probably needing some clever algebraic trick which I cannot seem to find.. 
 A: Since the partial derivatives are continuous at $(1,1)$, $f$ is differentiable there.Here's a proof of the fact  that $f'(1,1)(x,y)=3x+y$ from the definition. Let $\varphi(x,y)=3x+y$. Note that\begin{align}f(x,y)-f(1,1)-\varphi\bigl((x,y)-(1,1)\bigr)&=2xy+\frac xy-3x-y+1\\&=\left(2-\frac1y\right)(x-1)(y-1)+\frac{(y-1)^2}y.\end{align}Now, take $\varepsilon>0$. If $\bigl\lVert(x,y)\bigr\rVert<\frac14$, then $\lvert y-1\rvert<\frac14$ and so $\lvert y\rvert>\frac34$ and $\left\lvert2-\frac1y\right\rvert<\frac23$. So\begin{align}\left\lvert f(x,y)-f(1,1)-\varphi\bigl((x,y)-(1,1)\bigr)\right\rvert&=\left\lvert\left(2-\frac1y\right)(x-1)(y-1)+\frac{(y-1)^2}y\right\rvert\\&\leqslant\frac23\lvert x-1\rvert\lvert y-1\rvert+\frac43\lvert y-1\rvert^2.\end{align}So, take $\delta>0$ such that $\delta<\frac12$ and that $\delta<\frac\varepsilon2$. Then, if $\lVert(x,y)\rVert=r<\delta$, you have $\lvert x-1\rvert,\lvert y-1\rvert\leqslant r$ and\begin{align}\frac{\left\lvert f(x,y)-f(1,1)-\varphi\bigl((x,y)-(1,1)\bigr)\right\rvert}{\lVert(x,y)\rVert}&\leqslant\frac23r+\frac43r\\&=2r\\&<2\delta\\&<\varepsilon.\end{align}
