Prove that $\exists y\forall xf(x,y) \Rightarrow \forall x\exists yf(x,y)$
Assume the opposite:
statement 1 says: there is a y so that for all x'es f is true.
So statement 1 does not lead to statement 2. So if statement 1 is true, two is false. So
$\exists y\forall xf(x,y) \Rightarrow \neg\forall x\exists yf(x,y)$ ~ $\exists x\exists y\neg f(x,y)$
The last statement means (there is an x so that there is a y so that f is untrue).
However the last statement is equivalent to $\exists y\exists x\neg f(x,y)$. Which is in contradiction with statement 1.
Is this proof correct? I'm very unsure.
No it's not correct. Assuming the opposite of $p\Rightarrow q$ does not mean $p\Rightarrow \neg q$.
For example assume that you're about to prove that $x\in\mathbb N\Rightarrow x\in\mathbb Q$ (which is false) by assuming that $x\in\mathbb N\Rightarrow x\notin\mathbb Q$ (which is known to be false since $\mathbb N\subset\mathbb Q$).
In addition $\exists y\exists x\neg f(x,y)$ doesn't contradict $\exists y\forall x f(x,y)$, an example would be if $f(x,y) \Leftrightarrow y=0$. There surely exists $x$ and $y$ such that $\neg y=0$, but still there exists an $y$ such that $y=0$ for all $x$.