I'm working through Complex Variables - An Introduction.
Exercises 1.2.5 says that: Let $\gamma(\rho,\theta) = (\rho \cos \theta, \rho \sin \theta)$. Show that if $f$ is differentiable in $\mathbb{C}^{\star} = \mathbb{C} - \{ 0 \}$, then $$\frac{\partial (f \circ \gamma)}{\partial \rho} = \frac{1}{\rho} \Bigg ( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} \Bigg ) \circ \gamma.$$
If $f$ is differentiable, then $$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$$ then integrating gives us $$f = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}$$ I can't seem to go any further than here.
This is just the chain rule:
$$ \begin{align*} \frac{\partial \left( f \circ \gamma\right)}{\partial \rho} &= \frac{\partial f}{\partial x}\circ \gamma \cdot \frac{\partial \gamma_1}{\partial \rho} + \frac{\partial f}{\partial y}\circ \gamma \cdot \frac{\partial \gamma_2}{\partial \rho} \\\ &= \cos \theta \frac{\partial f}{\partial x} \circ \gamma+ \sin \theta \frac{\partial f}{\partial y} \circ \gamma \end{align*} $$
Then since $$ \cos \theta = \frac{x}{\rho}, \, \sin\theta =\frac{y}{\rho} $$ we get $$ \frac{\partial \left( f\circ \gamma \right)}{\partial \rho} = \frac{1}{\rho}\left( x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} \right) \circ \gamma $$