First, you should remember how to do this in single variable. For example, how would you show
$$ \lim_{x \to 2} 5x = 10 $$
What you want to show is that: For all $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x-2| < \delta$ then $ |5x - 10| < \epsilon $. This can be done as follows:
$$ |5x - 10| = 5|x-2| < 5\delta < \epsilon $$
and so by choosing $\delta = \dfrac{\epsilon}{5}$ we are good.
Now for your case, it's a two variables so you need to show that: for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $ 0 < \sqrt{(x-1)^2 + (y-0)^2} < \delta$ then $|ye^x - 0| < \epsilon $
First, you should remember how to do this in single variable. For example, how would you show
$$ \lim_{x \to 2} 5x = 10 $$
What you want to show is that: For all $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x-2| < \delta$ then $ |5x - 10| < \epsilon $. This can be done as follows:
$$ |5x - 10| = 5|x-2| < 5\delta < \epsilon $$
and so by choosing $\delta = \dfrac{\epsilon}{5}$ we are good.
Now for your case, it's a two variables so you need to show that: for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $ 0 < \sqrt{(x-1)^2 + (y-0)^2} < \delta$ then $|ye^x - 0| < \epsilon $
Also notice that: $y < \sqrt{(x-1)^2 + (y-0)^2} $
can you take it from here?