$f(x)$ this immediately!

From $f(x) + 5f\left(\frac{1}{1-x}\right) = \dfrac1x$ , we have:

$\begin{cases} x = 5: & f(5) + 5f\left(-\frac14\right) & = \dfrac15 & ... (1) \\ x = -\frac14: & f\left(-\frac14\right) + 5f\left(\frac45\right) & = - 4 & ... (2) \\ x = \frac45: &f\left(\frac45\right) + 5f\left(5\right) & = \dfrac{5}{4} & ... (3) \end{cases}$

$\begin{cases} (1)-5(2): \quad f(5) - 25f\left(\frac45\right) = \dfrac{101}5 \quad ... (4) \\ 25(3)+(4): \quad 126f(5) = \dfrac{1029}{20} \implies f(5) = \dfrac{49}{120} \end{cases}$

$\implies a + b = 49 + 120 = \boxed{169}$

$f(x)+5f(\frac{1}{1-x})=\frac{1}{x} ...(1)$

Let x = $\frac{1}{1-x}$

$f(\frac{1}{1-x})+5f(\frac{x-1}{x})=1-x ...(2)$

Let x = $\frac{x-1}{x}$

$f(\frac{x-1}{x})+5f(x)=\frac{x}{x-1} ..(3)$

$(1)- 5*(2):$ $f(x)-25f(\frac{x-1}{x})=\frac{5x^{2}-5x+1}{x} ...(4)$

$(4)+25*(3):$ $126f(x)=\frac{5x^{3}+15x^{2}+6x+1}{x(x-1)}$ $f(x)=\frac{5x^{3}+15x^{2}+6x+1}{126x(x-1)}$

Substituting x=5, $f(5)=\frac{49}{120}$

$a+b= 49+120= \boxed{169}$