Can I resubstitute polar coordinates after differentiation? let's consider $f: \mathbb R^2 \to \mathbb R, \quad (x,y) \mapsto xy$
We calcualte the partial-derivatives:
$\partial_x f=y$ 
$\partial_y f=y$
Now let's sitch to polar-coordinates, we get:
$f: (0,\infty)\times\mathbb R \to \mathbb R, \quad (r,\varphi) \mapsto r^2\sin \varphi \cos \varphi$
We calcualte the partial-derivatives:
$\partial_r f = 2r\sin\varphi\cos\varphi$ 
$\partial_\varphi f = r^2(-\cos^2\varphi+\sin^2\varphi)$
Now, If I calcualted the derivation in polra-coordinates, can I somehwo go back to eucledean coordinates? Just using $x=r\cos\varphi, y=r\sin\varphi$ won't do it, since if we take the partial-derivatives in polar coordinates we would get:
$2r\sin\varphi\cos\varphi=2y\cos\varphi$
$r^2\sin\varphi - r^2\cos^2\varphi = x^2-y^2$
I first expected to get $x$ and $y$ but that's bullshit since we aren't deriving in the same directions. So how could I now switch properly to eucleadean coordinates?
 A: Yes, you can translate partial derivatives between coordinate systems. However, there is a bit more work to it than there is to pure function values.
The reason for this is that the partial derivatives don't measure derivatives in the same direction. For instance, $\partial_r f$ measures how much $f$ changes as you move directly away from the origin, while $\partial_xf$ and $\partial_yf$ measure how much the function changes as you move horizontally and vertically. So if you take $\partial_rf$, and do the "naive" translation, the value you will get is the change in $f$ as you move away from the origin, expressed using cartesian coordinates.
So, what you have after your translation is a pair of functions $2y\frac{x}{\sqrt{x^2 + y^2}}$ (where I have removed the $\varphi$ completely) and $x^2 -y^2$. They tell you that if you're at the point $(x_0, y_0)$, and move some small distance $d$ directly away from the origin, then the function value will change by approximately $d\cdot 2y\frac{x}{\sqrt{x^2 + y^2}}$, and if you change the angle of the point (as in change $\varphi$) by a small amount $d$, then the function value changes by approximately $d(x^2 -y^2)$.
What you want is how much the function value changes if we move a short distance $d$ horizontally or vertically. I'll do the horizontal part here. Here we get that a change of $d$ horizontally becomes a change by
$$
r(x_0 + d, y_0) - r(x_0, y_0) = d\cdot \frac{r(x_0 + d, y_0) - r(x_0, y_0)}{d} \approx d\cdot\partial_x r(x_0, y_0)
$$
in the distance to the origin, and a small change of
$$
\varphi(x_0 + d, y_0) - \varphi(x_0, y_0) = d\cdot\frac{\varphi(x_0 + d, y_0) -\varphi(x_0, y_0) }{d}\approx d\cdot\varphi(x_0, y_0)
$$
in angle. So in total, a small horizontal change of $d$ results in some change in distance to the origin and some change in angle, each of which results in some change to the function value. Of course, these two changes will interfere with one another, but as $d$ gets smaller and smaller, they will interfere less and less. In the limit, we do actually get that
$$
\partial_x f(x_0, y_0) = \partial_rf(x_0, y_0)\cdot \partial_xr(x_0, y_0) +  \partial_\varphi f(x_0, y_0)\cdot \partial_x\varphi(x_0, y_0)
$$and similarily for $y$.
If you are familiar with linear algebra, we can leverage the concept of matrix multiplication to our advantage, and put the above result into a framework using what is known as the Jacobian matrix. Any (differentiable) change of coordinate systems comes with such a Jacobian matrix. This matrix has as entries the partial derivatives of the old coordinates with respect to the new ones. In our case, we have
$$
J = \begin{bmatrix}\partial_x r(x, y) & \partial_x\varphi(x, y)\\\partial_y r(x, y) & \partial_y \varphi(x, y)\end{bmatrix}
$$
This is a function $\Bbb R^2\to \Bbb R^{2\times 2}$, so each point gets its own matrix detailing exactly how the two coordinate grids relate to one another at that point. How it works is that if you have the partial derivatives with respect to $r, \varphi$ of some function at some point as a column vector $v$, then the partial derivatives with respect to $x, y$ at that point is just $Jv$.
For a specific example, consider your function at the point $p = (1, 1)$. At that point, the Jacobian for transforming from polar to cartesian is
$$
J_p = \begin{bmatrix}\frac{2\cdot 1}{2\sqrt{1^2 + 1^2}} & -\frac{1}{1^2 + 1^2}\\\frac{2\cdot 1}{2\sqrt{1^2 + 1^2}} & \frac{1}{1^2 + 1^2}\end{bmatrix} = \begin{bmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{2}\\\frac{1}{\sqrt{2}} & \frac{1}{2}\end{bmatrix}
$$
and the polar partial derivatives are $\partial_r f(1, 1) = \sqrt2, \partial_\varphi f(1, 1) = 0$. Thus $v_p = \left[\begin{smallmatrix}\sqrt2\\0\end{smallmatrix}\right]$, and
$$
\begin{bmatrix}\partial_x f(1, 1)\\\partial_y f(1, 1)\end{bmatrix} = J_pv_p = \begin{bmatrix}1\\1\end{bmatrix}
$$
This Jacobian framework also generalizes to Euclidean space (actually to smooth manifolds) of any dimension.
