lambda calculus beta reductions: ((((lambda f (lambda x ((f x) f))) (lambda y (lambda g (g (* y y))))) 2) (lambda a a))
$$
(\lambda f. \lambda x .f x f)
(\lambda y.\lambda g. g (* y y))
2
(\lambda a. a)
$$

The leftmost redex $(\lambda a.b)c$ is

$$(\lambda f. \lambda x .f x f)(\lambda y.\lambda g. g (* y y))$$

with


$a = f$
$b = \lambda x .f x f$
$c = (\lambda y.\lambda g. g (* y y))$


Don't forget to make sure that $x$ doesn't exist in $c$, because if so then doing an immediate beta reduction would change the meaning of the expression.  So the redex reduces to:

$$(\lambda x .(\lambda y.\lambda g. g (* y y)) x (\lambda y.\lambda g. g (* y y)))$$

Overall giving

$$
(\lambda x . (\lambda y.\lambda g. g (* y y)) x (\lambda y.\lambda g. g (* y y)))
2
(\lambda a. a)
$$

The next redex is $(\lambda y. \cdots)x$