Pretty Easy!!

$\begin{aligned} \lim_{t \to x} \frac {\sec x f(t) - f(x) \sec t}{t-x} & = \sec^2 x & \small \color{#3D99F6} \text{Let }t = x + \Delta x \\ \lim_{\Delta x \to 0} \frac {\sec x f(x+\Delta x) - f(x) \sec (x+\Delta x)}{\Delta x} & = \sec^2 x \\ \lim_{\Delta x \to 0} \frac {\sec x f(x+\Delta x) - \sec x f(x) + f(x) \sec x - f(x) \sec (x+\Delta x)}{\Delta x} & = \sec^2 x \\ \sec x \frac {df(x)}{dx} - f(x) \frac {d \sec x}{dx} & = \sec^2 x \\ \sec x f'(x) - f(x) \sec x \tan x & = \sec^2 x \\ f'(x) - f(x) \tan x & = \sec x \\ \implies f'(0) & = 1 \end{aligned}$

Since $f(0) = 1$ , $\dfrac {f'(0)}{f(0)} = \boxed 1$ .

$\displaystyle\lim_{t\to x}\dfrac {\sec(x)f(t) - f(x)\sec(t)}{t - x} = \sec^{2}(x)$

Now, applying L-H Rule ,we get(with respect to t)

$\dfrac {\sec(x)f'(x) - f(x)\sec(x)\tan(x)}{1 - 0} = \sec^{2}(x)$ .

Now, We know that,

$f(0) = 1$

Substituting, $x = 0$ , we get,

$\sec(0)f'(0) - f(0)\sec(0)\tan(0) = sec^{2}(0)$

$(1)f'(0) - (1)(1)(0) = (1)^{2}$

Then, $f'(0) = 1$

So, $\dfrac {f'(0)}{f(0)} = \dfrac {1}{1} = 1$

Therefore, $ANSWER=\boxed{1}$ .