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I know that x and y are the direction of basis in the space, but is there any condition for saying $\frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial r} = f_xcos(\theta)+f_ysin(\theta)$ ?
and why $\frac{\partial x}{\partial r}$ is equal to $cos(\theta)$? I can't intuitively justify it. why the derivation of $x$ to $r$ is equal to cosine?

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    Is that some kind of coordinate transformation between Cartesian and polar coordinates? – 2017-01-08
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    @user251257 yes, it is, but I can't get the details. – 2017-01-08
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    If this is the polar to cartesian coordinate transformation, then $$f(r,\theta)=(r\cos(\theta),r\sin(\theta))$$. So $x=r\cos(\theta)$ is linear in $r$. – 2017-01-08

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That is just because $x = r \cos \theta$ and $y = r\sin\theta$, hence $$ \def\p#1#2{\frac{\partial #1}{\partial #2}}\p xr = \cos\theta, \qquad \p yr = \sin \theta $$ Therefore, by the chain rule $$ \p fr = \p fx \p xr + \p fy \p yr = \p fx \cos\theta + \p fy \sin\theta $$

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    thanks for your answer, but why is it ok to write it as the sum of x and y? – 2017-01-08
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    Consider a point $p =(x,y)$. By definition of polar coordinates, $\theta$ is the angle between the positive $x$-axis and the half-line starting from $0$ through $p$. Hence, looking at the triangle $(0,0) -- (x,0) -- (x,y)$, we see that $x$ is the length of the side opposing $\theta$ and $r = (x^2 +y^2)^{1/2}$ is the length of the hypotenusis. Therefore $\cos\theta = \frac xr$, hence $x= r\cos\theta$: – 2017-01-08
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    thank you, but my problem is the "sum" why it is not something like $\frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}$. I'm sorry if that's an obvious question, but I couldn't find the details of this transformation by search. – 2017-01-08
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    That the chain rule. We do not just write it as this, we apply the chain rule, for the maps. – 2017-01-08