Multi-variable differentiation

Question

Let $\varphi , f: \mathbb{R}^2 \rightarrow \mathbb{R} $ differentiable functions, such that: $$\varphi (cx -af(x,y),cy-bf(x,y))=0, \text{ where } a,b,c \in \mathbb{R} \text{ and } a\frac{\partial \varphi}{\partial x} + b\frac{\partial \varphi}{\partial y} \neq 0$$ How do I show that $ a \frac{\partial f}{\partial x} + b \frac{\partial f}{\partial y} =c$?
I already played a bit with the function after using the chain rule but am not getting anywhere. Any advice is appreciated!

Answer 1

You need to assume $c\ne 0$ and that $\nabla\phi\ne 0$. (The $a$ and $b$ in your condition are not relevant.)

Write out the chain rule in its matrix form, and you see that it gives you $$(\nabla\phi)^\top\underbrace{\begin{bmatrix} c-a\frac{\partial f}{\partial x} & -a\frac{\partial f}{\partial y}\\-b\frac{\partial f}{\partial x} & c-b\frac{\partial f}{\partial y}\end{bmatrix}}_A = 0.$$ Since $\nabla\phi\ne 0$, the matrix $A$ must be singular, and so its determinant must vanish. With $c\ne 0$, this gives you your result.