When you take partial derivatives, what you are doing is to hold the other variable fixed. So for instance if $f$ is a function of $n$ variables, $f(x_1,x_2,\ldots,x_n)$, $\frac{\partial f}{\partial x_i}$ means you hold all the other $x_j$'s constant except $x_i$ and vary $x_i$ by $\delta x_i$ and find out what happens to $\delta f$.
We have
$x= r \cos(\theta)$,$y= r \sin(\theta)$, $r^2= x^2 + y^2$ and $\tan(\theta) = \frac{y}{x}$.
We are transforming from the $(x,y)$ space to $(r,\theta)$ space.
In the $(x,y)$ space, $x$ and $y$ are independent variables and In the $(r,\theta)$ space, $r$ and $\theta$ are independent variables.
$\frac{\partial x}{\partial r}$ means you are fixing $\theta$ and finding out how changing $r$ affects $x$.
So $\frac{\partial x}{\partial r} = \cos(\theta)$ since $\theta$ is fixed.
$\frac{\partial r}{\partial x}$ means you are fixing $y$ and finding out how changing $x$ affects $r$.
So $\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}$ since $y$ is fixed.
When you do $r = \frac{x}{\cos(\theta)}$ and argue that $\frac{\partial r}{\partial x} = \frac{1}{\cos (\theta)}$ you are not holding $y$ constant.
If you were to hold $y$ constant, then $\theta$ would change as well.
EDIT:
I am adding this in the hope that this will make it a bit more clear. If you want to use $r = \frac{x}{\cos(\theta)}$ and still derive it, you need to do as follows:
$r = \frac{x}{\cos(\theta)}$, $\delta r = \frac{\delta x}{\cos(\theta)} + \frac{-x}{\cos^2(\theta)} (-\sin(\theta)) \delta \theta$.
$\tan(\theta) = \frac{y}{x} \Rightarrow x \tan(\theta) = y$
$\delta x \tan(\theta) + x \sec^2(\theta) \delta \theta = 0$ (Since $y$ is held constant)
$x \delta \theta = - \frac{\delta x \tan(\theta)}{\sec^2{\theta}} = - \sin(\theta) \cos(\theta) \delta x$.
Plugging the above in the previous expression, we get
$\delta r = \frac{\delta x}{\cos(\theta)} + \frac{\sin(\theta)}{\cos^2(\theta)} (-\sin(\theta) \cos(\theta) \delta x) = \frac{1-\sin^2(\theta)}{\cos(\theta)} \delta x = \cos(\theta) \delta x$ and hence we get
$\frac{\partial r}{\partial x} = \cos(\theta) = \frac{x}{\sqrt{x^2+y^2}}$
Here is my proof:
$$ \frac{\partial}{\partial x}=\frac{\partial}{\partial r} \frac{\partial r}{\partial x} +\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}$$ so by applying the product rule
$$ \frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial r} \frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}\right)=
\frac{\partial^2}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2+
\frac{\partial}{\partial r}\frac{\partial^2 r}{\partial x^2} +
\frac{\partial^2}{\partial \theta^2}\left(\frac{\partial \theta}{\partial x}\right)^2+
\frac{\partial}{\partial \theta}\frac{\partial^2 \theta}{\partial x^2}$$
$$ +\frac{\partial^2 }{\partial r \partial \theta}\frac{\partial r}{\partial x} \frac{\partial \theta}{\partial x}$$
You get the same relation with $y$. Now you just have to calculate the derivatives of $r,\theta$ with respect to $x,y$. You have
$$ r=\sqrt{x^2+y^2},\ \theta=\arctan \frac{y}{x}$$.
What follows is a simple calculus exercise on derivatives. You just need to prove that
$$\left(\frac{\partial r}{\partial x}\right)^2+\left(\frac{\partial r}{\partial y}\right)^2=1, $$
etc.(the relations you need so that when you sum $\frac{\partial^2}{\partial x^2}$ and $\frac{\partial^2}{\partial y^2}$ you'll get the polar form of the laplacian).
Going on the lines you started, you shouldn't change $\sin \theta,\cos \theta$ in terms of $x,y$. Just calculate the next derivative with respect to $r,\theta$, using appropriately the formula of the partial derivative of composition of functions.
Best Answer
we have $$ u_x=f_rr_x, u_{xx}=f_rr_{xx}+r_x(f_{rr}r_x)\text{ similarly for } y, $$ $$ u_{xx}+u_{yy}=f_rr_{xx}+r_x(f_{rr}r_x)+f_rr_{yy}+r_y(f_{rr}r_y) =f_{rr}(r_x^2+r_y^2)+f_r(r_{xx}+r_{yy}). $$ with $$ r_x=\frac{x}{\sqrt{x^2+y^2}}, r_{xx}=\frac{y^2}{(x^2+y^2)^{3/2}} \text{ similarly for } y $$ we have $$ u_{xx}+u_{yy}=f_{rr}+\frac{1}{r}f_r. $$ of course if $f$ depends on $\theta$ it gets more complicated