When solving the Laplacian in polar coordinates, $x=r\cos\theta$ and $r^2=x^2+y^2$. When finding $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}$, the formula $r^2=x^2+y^2$, is used to find $\frac{\partial r}{\partial x}$:
$2r\frac{\partial r}{\partial x}=2x$
$\frac{\partial r}{\partial x}=\frac{x}{r}=\frac{r\cos\theta}{r}=\cos(\theta)$
Why, is it wrong to say: $r=\frac{x}{\cos\theta}$
$\frac{\partial r}{\partial x}=\frac{1}{\cos\theta}$
This sort of confusion arises because of our terrible notation for partial derivatives, which doesn't indicate which variables are being held constant. A good approach is to always indicate this explicitly until you've mastered partial derivatives to the point where you no longer get confused by suppressing this information in the notation.
In the present case, when you say you're looking for $\def\part#1#2{\frac{\partial#1}{\partial #2}}\part rx$, presumably you mean that you're looking for $\def\pt#1#2#3{\left.\part#1#2\right|_{#3}}\pt rxy$, with $y$ being held constant. When you write
$$ r=\frac x{\cos\theta}\;,\\ \part rx=\frac1{\cos\theta}\;, $$
that's only correctly differentiated if you're keeping $\theta$ constant while varying $x$. That is, you correctly calculated
$$ \pt rx\theta=\frac1{\cos\theta}\;. $$
The only problem is that that's not what you set out to find, since in general $\pt rx\theta\ne\pt rxy$.