It would help if someone can explain to me what I read in my book.
The equation is $x=rcos\theta$. We want to find $\frac{\partial r}{\partial x}$. One way to find it is to leave the equation as is and find $\frac{\partial x}{\partial r}$, which is $cos\theta$ and take the reciprocal of that to get $\frac{1}{cos\theta}$. The other way is to tweak the equation to be $r=\frac{x}{cos\theta}$ and find $\frac{\partial r}{\partial x}$ straight away and get $\frac{1}{cos\theta}$.
But then, here is where I get confused. The author says because we know $r=\sqrt{x^2+y^2}$, we can then find $\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=cos\theta$. The author explains the reason we are getting this answer is because here, instead of holding $\theta$ constant, we are pre-supposing $y$ to be constant and when $x$ varies, both $\theta$ and $r$ vary. Finally because this is not partial derivative by its definition, $cos\theta$ is rejected as a answer.
Note the following general principle: In an $n$-dimensional environment you can only talk about partial derivatives after a full set of $n$ independent variables (coordinate functions) has been specified.
In the case of the $(x,y)$-plane with additional "coordinate variables" $r$ and $\theta$ into the game you can consider, e.g., $x$ and $y$ as your "free" variables. Then any function $u$ defined in the plane, in particular $r$ and $\theta$, has well defined partial derivatives ${\partial u\over\partial x}$ and ${\partial u\over\partial y}$. One obtains $${\partial r\over\partial x}={2x\over\sqrt{x^2+y^2}}={x\over r},\qquad {\partial r\over\partial y}={2y\over\sqrt{x^2+y^2}}={y\over r}\ ,$$ and $${\partial\theta\over\partial x}={-y\over x^2+y^2},\qquad {\partial\theta\over\partial y}={x\over x^2+y^2}\ .$$ If you adopt $r$ and $\theta$ as your "free" variables then you obtain from $x=r\cos\theta$, $y=r\sin\theta$ that $${\partial x\over\partial r}=\cos\theta,\quad{\partial x\over\partial\theta}=-r\sin\theta,\quad\ldots\ .$$