I am trying to solve a problem:
Given that the function $f(x,y)$ satisfies the following equation: $$x\dfrac{df}{dy}+y\dfrac{df}{dx} = 0,$$ Prove or disprove: $$f(x,y) = f(\sqrt{x^2+t},\sqrt{\smash[b]{y^2+t}})$$ for every $t\geqslant0$, $x>0$, $y>0$.
Hint: prove that $\phi(t) = f(\sqrt{x^2+t},\sqrt{\smash[b]{y^2+t}})$ is a constant function.
I have done something that might solve that, but I am absolutely not sure if it is right. I marked $x(t) = \sqrt{x^2+t}$, $y(t) = \sqrt{y^2+t}$ and then I wrote this using the chain rule: $$\phi'(t) = \dfrac{df}{dx}\dfrac{dx}{dt}+\dfrac{df}{dy}\dfrac{dy}{dt}.$$ Is this correct? If it is correct, then I have managed to solve it, but I really doubt it is. If not, can someone tell me how can I prove $\phi$ is constant?
Thanks.
This is correct. In your case, the chain rule can be formally written as
$$ \begin{aligned} \phi'(t)&=\frac{\mathrm d\!}{\mathrm d t}\left(f\circ(x(t),y(t))\right)=\nabla f\cdot \left(\frac{\mathrm d x}{\mathrm d t},\frac{\mathrm d y}{\mathrm d t}\right)\\ &=\frac{\partial f}{\partial x}\frac{\mathrm d x}{\mathrm d t}+\frac{\partial f}{\partial y}\frac{\mathrm d y}{\mathrm d t}, \end{aligned} $$ which is exactly your formula (although you should remember to write $\partial f$ (\partial in LaTeX) instead of $\mathrm d f$ since later on these two symbols will have different meanings).