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.

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 $(a,b)$.
