Give a computation of the expression to normal form (Lambda calculus)
Usually, a series of applications $f\,x_1\,x_2 \dots$ means $\big(\big((f\,x_1)\,x_2\big)\,\ldots\big)$. We say that application is left associative.

According to this convention, there is an extra set of parentheses that is implied in the expression you write. Here I've made it explicit:
$$\big((\lambda f.\lambda g.\lambda x. f \,(g\,x))\,(\lambda u.u)\big)\,(\lambda v.v)$$

The expression $(\lambda f.\lambda g.\lambda x. f \,(g\,x))$ is a function and it is applied on input $(\lambda u.u)$. Reducing this will result to a function of the form $(\lambda g. \lambda x. \langle\text{body}\rangle)$, which in turn gets applied to $(\lambda v.v)$.