Remember to keep the symbols straight; there is no y in the actual problem.

Though I know what you mean. Acceleration is defined to be the second derivative of position with respect to time t, so we want to find s''(t).

First we find s'(t), which can be found by applying the chain rule, where the square root is the outer function and $\displaystyle t^3 + 1$ is the inner function. To keep things clear, I like to use exponents in place of square roots, so I would write it out:

$\displaystyle s(t) = (t^3 + 1)^\frac{1}{2}$

Explicitly, we can see that s(t) = f(g(t)) where $\displaystyle g(t) = t^3 + 1$ and $\displaystyle f(t) = t^{\frac{1}{2}}$. By the chain rule, s'(t) = f'(g(t))g'(t).

$\displaystyle s'(t) = \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}}(3t^2)$

Explicitly, we can see that s(t) = f(g(t)) where g(t) = t^3 + 1 and f(t) = t^(1/2)

In order to get the second derivative you will have to use the product rule in addition to the chain rule. Did you find your problem here?