Given relation between displacement and time

$x^2 = 1+t^2$

We need to find acceleration $a$ in terms of $x$, and compare it with

$a = x^{-n}$

Differentiate once to get velocity

Differentiating both sides with respect to $t$:

$2x\frac{dx}{dt} = 2t$

$x\frac{dx}{dt} = t$

Hence velocity,

$v = \frac{dx}{dt} = \frac{t}{x}$

Differentiate again to get acceleration

Since

$v = \frac{t}{x}$

acceleration is

$a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{t}{x}\right)$

Using quotient/product rule:

$a = \frac{x\cdot 1 - t\cdot \frac{dx}{dt}}{x^2}$

Substitute $\frac{dx}{dt} = \frac{t}{x}$:

$a = \frac{x - t\left(\frac{t}{x}\right)}{x^2}$

$a = \frac{x - \frac{t^2}{x}}{x^2}$

$a = \frac{x^2 - t^2}{x^3}$

Use the given relation $x^2 = 1+t^2$

$x^2 - t^2 = 1$

Therefore,

$a = \frac{1}{x^3} = x^{-3}$

Compare with $a = x^{-n}$

$x^{-n} = x^{-3}$

So,

$n=3$