If $f$ is differentiable in $(a,b)$ then $\frac{1}{f}$ is differentiable at $(a,b)$, provided $f(a,b)\neq0$ "Suppose that $f$ is a differentiable function at $(a,b)$. Prove that $\frac{1}{f}$ is differentiable in $(a,b)$, provided $f(a,b)\neq0$"
We were given the following definition of differentiability: a function is differentiable in $(a,b)$ if there exists a linear map $L$ such that $f(x,y)=f(a,b)+L_{(a,b)}\cdot(x-a,y-b)+r(x,y)$ with $\frac{r(x,y)}{|(x-a,y-b)|}\to0$. However, I still have no idea how to use this to prove the theorem. I guess you want to end up with something like $\frac{1}{f(x,y)}=\frac{1}{f(a,b)}+L_{(a,b)}\cdot(x-a,y-b)+r(x,y)$ with $\frac{r(x,y)}{|(x-a,y-b)|}\to0$, but I don't see how you would get there.
Please help. Thanks in advance.
 A: If f is differentiable then:
$\lim_\limits{(x,y)\to \mathbf (a,b)} \frac{f(\mathbf (x,y)) - f(a,b)}{|(x-a),(y-b)|}$ exists.
let $\mathbf x = (x,y), \mathbf a = (a,b)$
$\lim_\limits{\mathbf x\to\mathbf a} \frac{\frac {1}{f(\mathbf x)} - \frac{1}{f(\mathbf a)}}{|\mathbf x-\mathbf a|}\\
\lim_\limits{\mathbf x\to \mathbf a} \frac{1}{f(\mathbf x)f(\mathbf a)} \frac{{f(\mathbf a) - f(\mathbf x)}}{|\mathbf x-\mathbf a|}\\$
If $f(\mathbf a) \ne 0$ then the limit above exists.
A: To be able to prove differentiability, you first need to guess what the linear map is. Usually one does this by using differentiation rules (derivative of a product, chain rule) that have appropriate extensions to the multivariable setting. But let us work this directly.
A decent rule of thumb is to mimic the one-variable rules. So we may guess, from $$ \left(\frac1f\right)'(t)=-\frac{f'(t)}{f(t)^2},$$ that the linear map $M$ for $1/f$ should be 
$$
M_{(a,b)}(x,y)=-\frac{L_{(a,b)}(x,y)}{f(a,b)^2}.
$$
Let us give it a try: because $f(a,b)\ne0$ and $f$ is continuous (being differentiable), there exist $c,d>0$ with $d\geq|f(a+x,b+y)|\geq c$ for $x,y$ small enough. Then
\begin{align}
\frac{\left|\displaystyle\frac1{f(a+x,b+y)}-\frac1{f(a,b)}+\frac{L_{(a,b)}(x,y)}{f(a,b)^2}\right|}{(x^2+y^2)^{1/2}}
&=
\frac{\left|\displaystyle{f(a,b)^2}{}-{f(a+x,b+y)f(a,b)}{}+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{(x^2+y^2)^{1/2}|f(a+x,b+y)|f(a,b)^2}\\ \ \\
&\leq
\frac{\left|\displaystyle{f(a,b)^2}{}-{f(a+x,b+y)f(a,b)}{}+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle{f(a,b)}(f(a,b)-{f(a+x,b+y))}{}+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle{f(a,b)}(-L_{(a,b)}(x,y)-r(x,y))+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle(f(a+x,b+y)-f(a,b))\,L_{(a,b)}(x,y)-f(a,b)\,r(x,y))\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle(L_{(a,b)}(x,y)+r(x,y))\,L_{(a,b)}(x,y)-f(a,b)\,r(x,y))\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&\leq
\frac{\displaystyle(L_{(a,b)}(x,y)^2}{c^3(x^2+y^2)^{1/2}}+\frac{|r(x,y))\,L_{(a,b)}(x,y)|}{c^3(x^2+y^2)^{1/2}}+\frac{|f(a,b)\,r(x,y))|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&\leq
\frac{\displaystyle|L_{(a,b)}|^2(x^2+y^2)}{c^3(x^2+y^2)^{1/2}}+\frac{|r(x,y)|\,|L_{(a,b)}|^2(x^2+y^2)}{c^3(x^2+y^2)^{1/2}}+\frac{d\,|r(x,y))|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\displaystyle|L_{(a,b)}|^2(x^2+y^2)^{1/2}}{c^3}+\frac{|r(x,y)|\,|L_{(a,b)}|^2(x^2+y^2)^{1/2}}{c^3}+\frac{d\,|r(x,y))|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
\end{align}
and now it is clear that the three terms go to zero as $(x,y)\to0$, so $1/f$ is differentiable at $(a,b)$.
