Derivative of $f(3x+1,3x-1)=4$ This exercise asks me to take the derivative of
$$f(3x+1,3x-1)=4$$
where this equality is said to be valid for all $x$.
The exercise specifically asks me to prove that
$$\frac{∂}{∂x}f(3x+1,3x-1)=-\frac{∂}{∂x}f(3x+1,3x-1)$$
The first thing I though was to apply the partial derivative operator to the both sides of the function:
$$\frac{∂}{∂x}f(3x+1,3x-1)=\frac{∂}{∂x}4 \implies \frac{∂}{∂x}f(3x+1,3x-1) = 0 \implies \\ \frac{∂}{∂x}f(3x+1,3x-1) = - \frac{∂}{∂x}f(3x+1,3x-1)$$
but the exercise uses the chain rule, so I'm assuming that this can't be made. Could someone clarify for me what am I doing wrong here?
 A: Here, we have a function $f$ of two variables, say $s$ and $t$, and in turn $s$ and $t$ are functions of a single variable $x$.  That is we have $f(s,t)$ and $s=3x+1$ and $t=3x-1$.
So, we introduce a new function, $g(x)=f(s(x),t(x))=f(3x+1,3x-1)$.  We are given that $g(x)=4$ for all $x$.  And this implies that $g'(x)=0$ for all $x$.  So, from the chain-rule, we have
$$\begin{align}
g'(x)&=\left.\frac{\partial f(s,t)}{\partial s}\right|_{s=3x+1,t=3x-1}\frac{ds}{dx}+\left.\frac{\partial f(s,t)}{\partial t}\right|_{s=3x+1,t=3x-1}\frac{dt}{dx}\\\\
&=3\left.\frac{\partial f(s,t)}{\partial s}\right|_{s=3x+1,t=3x-1}+3\left.\frac{\partial f(s,t)}{\partial t}\right|_{s=3x+1,t=3x-1}\\\\
&=0
\end{align}$$
from which we see
$$\left.\frac{\partial f(s,t)}{\partial s}\right|_{s=3x+1,t=3x-1}=-\left.\frac{\partial f(s,t)}{\partial t}\right|_{s=3x+1,t=3x-1}$$
A: Let $I \subset \mathbb{R}$ be open; let $\varphi: x \mapsto (3x+1, 3x-1)$ on $I$; let $f\circ \varphi (x) = 4$ on $I$; and let
$g: x \mapsto \varphi(x) \mapsto f\circ \varphi(x) = 4$ on $I$. Then
$g'(x) = 0 = \big( D_{1}f(3x+1,3x-1), D_{2}f(3x+1, 3x-1) \big) \cdot (3,3)$ on $I$,
so
$$
D_{1}f(3x+1, 3x-1) = -D_{2}f(3x+1, 3x-1)
$$
on $I$.
