Differentiable

To solve this, we must know the first principle of differentiation according to which we have $\displaystyle \lim_{h\to 0}(\frac{f(x+h)-f(x)}{h})=f'(x)$ . Now, we have in this question that $f(x)$ is a function differentiable at $x=1$ with $f'(1)=\frac{1}{5}$ . So ----

$\displaystyle \lim_{x\to 1}(\frac{x^3-1}{f(x)-f(1)})$

$=\large \displaystyle \lim_{x\to 1}(\frac{\frac{x^3-1}{x-1}}{\frac{f(x)-f(1)}{x-1}})$ [dividing both numerator and denominator by $(x-1)$ ]

$=\frac{\displaystyle \lim_{x\to 1}(\frac{x^3-1}{x-1})}{\displaystyle \lim_{x-1\to 0}(\frac{f(x)-f(1)}{x-1})}$ [since $x\to 1 \implies x-1\to 0$ ]

$=\frac{\displaystyle \lim_{x\to 1}(\frac{(x-1)(x^2+x+1)}{(x-1)})}{\displaystyle \lim_{x-1\to 0}(\frac{f(1+(x-1))-f(1)}{x-1})}$ [since $a^3-b^3=(a-b)(a^2+ab+b^2)$ ]

$=\frac{\displaystyle \lim_{x\to 1}(x^2+x+1)}{f'(1)}$ [from first principle taking $x=1,h=(x-1)$ ]

$=\frac{1^2+1+1}{\frac{1}{5}} = \frac{1+1+1}{\frac{1}{5}}=\frac{3}{\frac{1}{5}}=3\times 5 =\boxed{15}$