I want to show that the function $e^{x+2y}$ is differentiable in the point $(2,2)$ by using the definition.
Definition: A multivariable function is differentiable in a point a if there exist constants $A_{1} + \dotsb + A_{n}$ and a function $\varrho(h)$ such that $$f(a+h)-f(a)=A_{1}h_{1}+ \dotsb+ A_{n}h_{n}+ |h|\varrho(h)$$ and $\varrho(h)\to 0$ as $h \to (0, 0)$.
I find the expression for $\varrho(h)$ to be the following:
$$\frac{e^{h+2k}-1-h-2k}{\sqrt{h^2+k^2}}$$
I write the function in terms of polar coordinates ($x= r \cos \theta, y =r \sin\theta$) and obtain:
$$\frac{e^{r \cos\theta +2r\sin\theta}-1-r \cos \theta-2r\sin\theta}{r}$$
Using the squeeze theorem I obtain:
\begin{align} \left|\frac{e^{r \cos \theta +2r\sin\theta}-1-r\cos \theta-2r\sin\theta}{r}\right|&\leq \left|\frac{e^{r+2r}+r+2r}{r} \right|\\ &= \frac{e^{r+2r}}{r} + \frac{r}{r}+ \frac{2r}{r}\\ &= \frac{e^{r+2r}}{r} +3 \end{align}
The limit of this as $r \to 0$ is $3$, but the answer should be $0$. I don't know where I went wrong.
Any help would be much appreciated!
Firstly your expression for $\rho(h)$ is not correct since it is missing a factor $e^6$. By your definition
$$ \rho(h) = \frac{f(a+h)-f(a)-A_1h_1-A_2h_2}{\vert h \vert } = \frac{e^{(2+h_1)+2(2+h_2)} - e^6- A_1h_1-A_2h_2}{\vert h \vert}$$
where the constants are $A_1 = \frac{\partial f }{\partial x}(2,2) = e^6$ and $A_2 = \frac{\partial f}{\partial y}f(2,2) = 2e^6.$ Hence
$$ \rho(h) = e^6\frac{e^{h_1+2h_2}-1-h_1-2h_2}{\sqrt{h_1^2+h_2^2}}.$$
Regardless we still need to compute the limit as $h \to 0 $ of
$$\frac{e^{h_1+2h_2}-1-h_1-2h_2}{\sqrt{h_1^2+h_2^2}}$$
as you did.
Secondly $\frac{e^{r+2r}}{r} + 3$ does not converge to $3$ since $e^{3r}/r$ diverges as $r \to 0$. Even if the limit was $3$ the squeeze theorem does not allow you to conclude that
$$ \lim_{(x,y) \to 0} \frac{e^{x+2y} -1 -x -2y}{\sqrt{x^2 + y^2}} = 3.$$
To use the squeeze theorem here you need a bound which converges to $0$.
Here's a solution to the problem.
Remember that $P(x) = 1 + x$ is the first order Taylor polynomial for $e^x$ at $0$. Hence by definition we have
$$ \lim_{x \to 0}\frac{e^x - P(x)}{x} = 0$$ i.e.
$$ \forall \epsilon , \exists \delta : \vert x \vert \leq \delta \implies \vert e^x - P(x)\vert \leq \epsilon \vert x \vert $$
Notice that
$$ \left\vert\frac{e^{x+2y} -1 - x -2y}{\sqrt{x^2+y^2}} \right\vert= \left\vert \frac{e^{x+2y} - P(x+2y)}{\sqrt{x^2+y^2}}\right\vert.$$
Let $\epsilon > 0$ and take $\vert (x,y) \vert < \tilde \delta$ with $\tilde \delta$ small enough to guaranty that $\vert x+ 2y\vert< \delta$ and we have
$$ \left\vert\frac{e^{x+2y} -1 - x -2y}{\sqrt{x^2+y^2}} \right\vert \leq \epsilon \frac{\vert x+2y\vert }{\sqrt{x^2+y^2}} \leq \epsilon \frac{3r}{r} = 3\epsilon$$
Therefore by definition we have that
$$ \lim_{(x,y) \to 0} \frac{e^{x+2y}-1 -x-2y}{\sqrt{x^2+y^2}} = 0$$
and $e^{x+2y}$ is differentiable at $(2,2)$