How to prove " $¬\forall x P(x)$ I have a step but can't figure out the rest. I have been trying to understand for hours and the slides don't help. I know that since I have "not P" that there is a case where not All(x) has P... but how do I show this logically?

1. $\forall x (P(x) → Q(x))$         Given
2. $¬Q(x)$             Given
3. $¬P(x)$             Modus Tollens using (1) and (2)
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First, you want to instantiate your quantified statement with a witness, say $x$: 
So from $(1)$ we get $$\;P(x) \rightarrow Q(x) \tag{$1\dagger$}$$
Then from $(1\dagger)$ with $(2)$ $\lnot Q(x)$, by modus tollens, you can correctly  infer $(3)$: $\lnot P(x)$.
So, from $(3)$ you can affirm the existence of an $x$ such that $\lnot P(x)$ holds: $\quad\exists x \lnot P(x)$
Then recall that, by DeMorgan's for quantifiers,$$\underbrace{\exists x \lnot P(x) \quad \equiv \quad \lnot \forall x P(x)}_{\text{these statements are equivalent}}$$
