How to find $f'(e^2)$ given that $f(x) + \ln (x^2 + f(x)^2) = 4$? So, as the title says, I'm looking to find $f'(e^2)$, being How can I find $f'(e^2)$ being $y = f(x)$ in $y + \ln (x^2 + y^2) = 4$. I found this interesting exercise in a peruan book, and I think that it'd be a great way to share its mechanics.
Firstly, I did the $f(x)$ equation:
\begin{align*}
y + \ln(x^2 + y^2) &= 4\\
f(x) + \ln[x + f(x)^2] &= 4\\
f(e²) + \ln[e^2 + f(e^2)^2] &= 4
\end{align*}
Then, the $f'(x)$ part:
\begin{align*}
[y + \ln(x^2 + y^2)]' &= [4]'\\
y' + \displaystyle\frac{1}{x^2 + y^2} \cdot (2x + 2yy') &= 0\\
y' &= -\displaystyle\frac{2x + 2yy'}{x^2 + y^2}\\
y' \cdot (x^2 + y^2) &= –(2x + 2yy')\\
y' \cdot (x^2 + y^2) &= –2x – 2yy'\\
y' \cdot (x^2 + y^2) + 2yy' &= –2x \\
y' \cdot (x^2 + 2y + y²)  &= –2x \\
y' &= –\displaystyle\frac{2x}{x^2 + 2y + y^2}\\
f'(x) &= –\displaystyle\frac{2x}{x^2 + 2f(x) + f(x)^2}\\
f '(e^2) &= –\displaystyle\frac{2e^2}{e^4 + 2f(e^2) + f(e^2)^2²}
\end{align*}
Couldn't go on from here.

So firstly we start differentiating and then look for corresponding value of the function.
$$y=4-\ln (x^2+y^2)$$
$$y'=-\frac{1}{x^2+y^2} ( 2x+2yy')$$
Simplifying gives
$$y'=-\frac{2x}{x^2+y^2+2y}$$
Evaluating at $x=e^2$ you'll be needing $f(e^2)$.
So try to substitute this in what is given and find y that is
$$y=4-\ln(e^4+y^2)$$
Now we substitute for few values and observe how the function behaves and substituting $y=0$ perfectly fits the condition as then you get $4-4=0$ on rhs.
Put this back in our calculated derivative to get
$$y'(e^2)=-\frac{2e^2}{e^4}=-\frac{2}{e^2}$$
