Consider $f:\Omega\subset\mathbb{R^2}\longrightarrow\mathbb{R}$ with $\Omega$ open and $f\in\mathcal{C^1(\Omega)}$.
Now we define another function $u:\mathbb{R^3}\longrightarrow\mathbb{R}$ defined as $u(x,y,z)=x^4\cdot f(y/x,z/x)$
I'm asked to prove that $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=4u$
My try:
Calculating those partial derivatives by using the product rule, multiplying by the respective variables and adding up, I end up with:
$4x^4f(y/x,z/x)+x^5\frac{\partial}{\partial x}f(y/x,z/x)+x^4y\frac{\partial}{\partial y}f(y/x,z/x)+x^4z\frac{\partial}{\partial z}f(y/x,z/x)$
Since the first term is already $4u$, for the equality to hold, I'd have to prove that:
$x^5\frac{\partial}{\partial x}f(y/x,z/x)+x^4y\frac{\partial}{\partial y}f(y/x,z/x)+x^4z\frac{\partial}{\partial z}f(y/x,z/x)=0$
But I have no idea how to prove such thing, as I don't know how to calculate those partial derivatives, there must be some chain rule to be applied here, but I'm unsure.
You need to apply the chain rule for several variables and everything will magically disappear: $$ \frac{\partial}{\partial x}f(\phi(x,y,z),\psi(x,y,z))= \frac{\partial f}{\partial\phi}\cdot\frac{\partial\phi}{\partial x}+ \frac{\partial f}{\partial\psi}\cdot\frac{\partial\psi}{\partial x} \quad\text{etc (for $y$ and $z$)} $$