For any transformation from Rectangular (Cartesian) to spherical coordinates in three dimensions, we use the transformation equation
$x = r\cos(\theta)\sin(\phi)$
among others, where $\theta$ is the azimuth and $\phi$ is the polar angle.
Now,
$\cfrac{\partial x}{\partial r} = \cos(\theta)\sin(\phi) = \cfrac{x}{r} \tag{1}$.
However, $r = \sqrt{x^2 + y^2 + z^2}$ so
$\cfrac{\partial r}{\partial x} = \cfrac{x}{\sqrt{x^2 + y^2 + z^2}} = \cfrac{x}{r} \tag{2}$.
What's wrong with the above?
Absolutely nothing, except that it is recommended not to mix different coordinate systems in the same expression.
I suspect you are concerned that this doesn't match up with $ \frac{dy}{dx} = \left( \frac{dx}{dy} \right)^{-1} $. This rule only works for total derivatives: for partial derivatives, the relationship between them depends on what is being held constant when you take the derivative.
Consider, for example, the change of coordinates $x = u$, $y=u+v$, so $u=x$ and $v=y-x$. You might think that $\partial f/\partial u$ is the same as $\partial f/\partial x$, but it isn't: we find $$ \frac{\partial f}{\partial u} = \frac{\partial x}{\partial u}\frac{\partial f}{\partial x} + \frac{\partial y}{\partial u} \frac{\partial f}{\partial y} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}, $$ so even though $u$ and $x$ look the same, holding the other coordinate constant in the differentiation gives different results (draw the lines $y=\text{const.}$ and $v=\text{const.}$ to see this clearly). We have in this case $$ \frac{\partial u}{\partial x} = 1, \quad \frac{\partial v}{\partial x} = -1 \\ \frac{\partial u}{\partial y} = 0, \quad \frac{\partial v}{\partial y} = 1, $$ but $$ \frac{\partial x}{\partial u} = 1, \quad \frac{\partial y}{\partial u} = 1 \\ \frac{\partial x}{\partial v} = 0, \quad \frac{\partial y}{\partial v} = 1, $$ so we see that we have a different reciprocal relation: the matrices of partial derivatives are inverses, or if we write $x=x_1$, $x=x_2$, $u=u_1$, $v=u_2$ $$ \sum_{j=1}^2 \frac{\partial x_i}{\partial u_j} \frac{\partial u_j}{\partial x_k} = \delta_{ik}. $$ This relation tells us very little about the individual partial derivatives.