(transcribed from a pic; couldn't find the source)
If a point $P(x,y)$ lies on the curve $y=f(x)$ such that the following limit exists: $$ \lim_{(x,y)\to(1,2)}\frac{\arctan(x)+\arctan(y^{-1})-\arctan(3)}{(x-1)(y-2)} \cdot\arcsin(y-2), $$then find $$ 10 \lim_{x\to 1/3}\frac{f^{-1}(x)}{3x-1} $$
I tried solving this question and came up with the following approach:
Apply L'Hopital's Rule (LHR): $$ =\lim_{(x,y)\to(1,2)} \frac{\frac{1}{1+x^2}dx+\frac{1}{1+1/y^2}\cdot (-y^{-2})dy}{dx} $$ $$ =\lim_{(x,y)\to(1,2)} \frac{1}{1+x^2}-\frac{1}{1+y^2}\cdot \frac{dy}{dx} = \frac{1}{2}-\frac{1}{5}f'(1) $$Since this limit exists, $f'(1)\ne \infty$. Now apply LHR to the limit in question: $$ 10\lim_{x\to 1/3} \frac{f^{-1}(x)}{3x-1} $$ $$ = \frac{10}{3} \lim_{x\to1/3} \frac{1}{f'\left(f^{-1}(x)\right)} $$Since $f(0)=1/3$, this becomes $$ \frac{10}{3} \frac{1}{f'(0)} $$Now we can let $f$ be any function, since it's not mentioned in the question. In particular, the linear function interpolating $(0,1/3)$ and $(1,2)$ works, which gives $f'(0)=5/3$. Then the given limit is $2$.
However the answer given is 3. Although I am doubtful of my solution I don’t know what exactly is wrong with it. Also, what would be the correct way to do it?