Finding the limit of $f(f(x))$ type problem Let $f$ be a differentiable function and equation
of normal to the graph of $y = f(x)$ at $x = 3$ is
$3y = x + 18$. If $L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) - f\left( {{{\left( {3 + f\left( 3 \right) - x - 6} \right)}^2}} \right)}}{{{{\sin }^2}\left( {x - 1} \right)}}$ then find the value of L.
My approach is as follow
$y = \frac{{x + 18}}{3} \Rightarrow f\left( x \right) = \frac{{x + 18}}{3}$
$f'\left( x \right) = \frac{1}{3}$
Confused after applyying L.Hosp rule
$L = \mathop {\lim }\limits_{x \to 1} \frac{{f'\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) \times 2\left( {4{{\tan }^{ - 1}}x - \pi } \right) \times \frac{4}{{1 + {x^2}}} - f'\left( {{{\left( {3 + f\left( 3 \right) - x - 6} \right)}^2}} \right) \times 2\left( {3 + f\left( 3 \right) - x - 6} \right) \times  - 1}}{{\sin 2\left( {x - 1} \right)}}$

Hint:
Normal $3y=x+18$ gives us $y=7$ for $x=3$ so it intersects the curve at $(3,7)$ so $f(3)=7$ and the slope of the normal is $\frac{1}{3}$, so $f'(3)=-3$ (normal being perpendicular to the tangent etc.).
