You cannot in general. To take a very simple example, consider $f(x,y) = \sin(xy)$, and $y'=0$. Then we have:
$$\lim_{y\to 0}\>\sup_{x\in\mathbb{R}}\sin(xy) = 1.$$
(Remember that if we take the limit as $y\to 0$, then we do not consider $y=0$, so $y\neq 0$ in $\sin(xy)$).
But
$$\sup_{x\in\mathbb{R}}\>\lim_{y\to 0}\sin(xy) = \sup_{x\in\mathbb{R}}\ 0 = 0.$$
However, if the limit of $f(x,y)$ as $y\to y'$ exists for every $x$, and the limit of the suprema exist, then you get one inequality:
$$\sup_{x\in\mathbb{R}}\>\lim_{y\to y'}f(x,y) \leq \lim_{y\to y'}\>\sup_{x\in\mathbb{R}} f(x,y).$$
To see this, note that for each fixed $x_0\in\mathbb{R}$, $f(x_0,y) \leq \sup\limits_{x\in\mathbb{R}}f(x,y)$, so taking limits we have
$$\lim_{y\to y'} f(x_0,y) \leq \lim_{y\to y'}\>\sup_{x\in\mathbb{R}}f(x,y).$$
Since this holds for each $x_0$, the supremum also satisfies the inequality, so
$$\sup_{x\in\mathbb{R}}\>\lim_{y\to y'}f(x,y) \leq \lim_{y\to y'}\>\sup_{x\in\mathbb{R}}f(x,y).$$