How to say that $f(x,y)$ is true for all x but only some y. I have to translate an English sentence into logical statements using quantifiers, and I think I'm on the right track but something doesn't feel right about this. I'm not really sure how I should take "as long as y satisfies..." into consideration.

For all $\epsilon>0$, for all real numbers $x$, one an find a real number $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ as long as $y$ satisfies $|y-x|<\delta$.

My proposed solution follows:
$$\forall\epsilon>0,\ \forall x\in\mathbb{R},\ \exists\delta>0,\ \exists y\ni |y-x|<\delta\Longrightarrow|f(x)-f(y)|<\epsilon$$
And as a secondary part of the question, I have to negate that statement:
$$\exists\epsilon>0,\ \exists x\in\mathbb{R},\ \forall\delta>0,\ \forall y,\ |y-x|<\delta\wedge |f(x)-f(y)|\geq\epsilon$$
 A: Let me start by saying that what you have on the grey box is not a statement. The variable $y$ isn't quantified. That formula is devoided of meaning.
Presumably the intent is that $y$ is universally quantified and its quantifier should be the inner-most one, so you get the definition of continuous function.  Thus your statement is:
$$\forall x\in \Bbb R\forall \varepsilon >0 \exists \delta >0\color{red}{\forall }y\in \Bbb R\left(|y-x|<\delta\implies |f(x)-f(y)|<\varepsilon\right)$$
and negating you get
$$\exists x\in \Bbb R \exists \varepsilon>0\forall \delta >0\color{red}\exists y\in \Bbb R\left(|y-x|<\delta \land |f(x)-f(y)|\ge \varepsilon\right).$$
A: "As long as" signifies the antecedent of a conditional (the "if" part). With a little switch in notation to make it cleaner, you could write
$$\forall\epsilon>0,\ \forall x\in\mathbb{R},\ \exists\delta>0,\ \forall y \in \mathbb{R}\;\; [y \in (x-\delta, x+\delta) \Longrightarrow |f(x)-f(y)|<\epsilon].$$
Let me know if you run into trouble negating that.
