$f^{-1}(x)$ Does not means $\frac{1}{f(x)}$

$g(x)=f^{ -1 }\left( x \right) \\ \\ f^{ -1 }\left( f\left( x \right) \right) =x\\ \\ g(f(x))=x$

Differentiating

$g^{ \prime }\left( f(x) \right) *f^{ \prime }\left( x \right) =1$

$g^{ \prime }\left( f(x) \right) =\frac { 1 }{ f^{ \prime }\left( x \right) } \\$ ....... $(1)$

Also $f(1)=5$ . u can check that by equation $f(x)=5$

On again differentiating $(1)$

$g^{ \prime \prime }\left( f(x) \right) =\frac { -f^{ \prime \prime }\left( x \right) }{ { (f^{ \prime }\left( x \right) })^{ 3 } }$

$f^{ \prime }\left( x \right) =3{ x }^{ 2 }+3\quad and\quad f^{ \prime \prime }\left( x \right) =6x$

$g^{ \prime \prime }(f(1))=\frac { -f^{ \prime \prime }\left( 1 \right) }{ { (f^{ \prime }\left( 1 \right) })^{ 3 } }$

$g^{ \prime \prime }(5)=\frac { -6 }{ { (6 })^{ 3 } }$

$\boxed{g^{ \prime \prime }(5)=\frac { -1 }{ 36 } }$

Do upvote

We note that since $f(1) = 5$ we have $g(5) = 1$ . Also, since $g(x)$ is the inverse of $f(x)$ it must satisfy: $x = g(x)^3 + 3g(x) + 1$ Using implicit differentiation, solving for $g^\prime(x)$ , and then evaluating at $x = 5$ gives us: $\begin{aligned} 1 &= 3 g^\prime(x) g(x)^2 + 3 g^\prime(x) \\ g^\prime(x) &= \frac{1}{3\left( g(x)^2 + 1 \right)} \\ g^\prime(5) &= \frac{1}{6} \end{aligned}$ Using implicit differentiation again, solving for $g^{\prime\prime}(x)$ , and then evaluating at $x = 5$ gives us: $\begin{aligned} 0 &= 6 \left( g^\prime(x) \right)^2 g(x) + 3 g^{\prime\prime}(x) g(x)^2 + 3 g^{\prime\prime}(x) \\ g^{\prime\prime}(x) &= -\frac{2 \left( g^\prime(x) \right)^2 g(x)}{g(x)^2 + 1} \\ g^{\prime\prime}(5) &= -\frac{1}{36} \end{aligned}$ Therefore the solution is $a + b = \boxed{37}$ .

Conceptual way can sometime be an alternative to genuine way:

1
2
3

4.999994    0.999999    0.166666750019786           
5   1   0.166666583313631   -0.027767842013975  -   1/36    -0.027777777777778
5.000006    1.000001    0.166666416706495           

We must have consciousness in mind to apply the above method.

a + b = 1 + 36 = 37

Answer: $\boxed{37}$