$f$ is assumed to be continuous only at $(x_{0}, y_{0})$, not at $(x, y_0)$ for an arbitrarily chosen $x$.
Recall that $\limsup_{x \to x_{0}}f(x) := \inf_{r > 0}\sup_{|x - x_{0}| < r}f(x)$ and $\liminf_{x \to x_{0}}f(x) := \sup_{r > 0}\inf_{|x - x_{0}| < r}f(x)$.
Note that the exact choice of metric is irrelevant here since talking about continuity in $\mathbb{R}^{n}$, it suffices to know that the distance is determined by the quantity of the form $|x_{i} - y_{i}|$. We give $\mathbb{R}^{2}$ the Euclidean metric, say.
By condition, $\forall \varepsilon > 0$, $\exists \delta > 0$, such that when $\sqrt{(x - x_{0})^{2} - (y - y_{0})^{2}} < \delta$, $|f(x, y) - f(x_{0}, y_{0})| < \varepsilon$.
Fix some $x$ with $|x - x_{0}| < \frac{\delta}{2}$. $\forall |y - y_{0}| < \frac{\delta}{2}$, $\sqrt{(x - x_{0})^{2} - (y - y_{0})^{2}} < \delta$. Hence by definition, $f(x_{0}, y_{0}) - \varepsilon \leq \sup_{|y - y_{0}|< r}f(x, y) \leq f(x_{0}, y_{0}) + \varepsilon$, when $r < \frac{\delta}{2}$. Taking the infimum over $r > 0$, we then get: $f(x_{0}, y_{0}) - \varepsilon \leq \inf_{r > 0}\sup_{|y - y_{0}| < r}f(x, y) \leq f(x_{0}, y_{0}) + \varepsilon$. In particular, we have:
$\forall |x - x_{0}| < \frac{\delta}{2}$, $|\limsup_{y \to y_{0}}f(x, y) - f(x_{0}, y_{0})| < \varepsilon$. ie, $\lim_{x\to x_0}\limsup_{y\to y_0}f(x,y) = f(x_{0}, y_{0})$.
A similar argument shows that $\lim_{y\to y_0}\limsup_{x\to x_0}f(x,y)=f(x_0,y_0)$.
Let $(a_n)$ be a sequence converging to $a$ with $a_n >a$, say $a_n=a+\frac 1 n$. Then, by MVT, $f(a_n)-f(a_m)=f'(t)(a_n-a_m)$ for some $t$ between $a_n$ an $a_m$. Let $l =\lim_{x \to a} f'(x) $. There exists $\delta >0$ such that $|f'(x)|<l+1$ if $x>a,|x-a| <\delta$. If $n$ and $m$ are sufficiently large then $|t-a| <\delta$ so $|f(a_n)-f(a_m)| <(l+1)|a_n-a_m| \to 0$. Hence, $f(a_n)$ is Cauchy sequence. Let $L=\lim f(a_n)$. If $(x_n)$ is any sequence converging to $a$ from the right we can apply MVT again to show that $f(x_n)-f(a_n) \to 0$. Hence, $f(x_n) \to L$. It follows that $f(x) \to L$ as $ x \to a$.
Best Answer
This is not true. Consider the function $$f(x)=\begin{cases}1&\text{if }x\geq0\\0&\text{if }x<0\end{cases}.$$ Then $\lim_{x\to0}f(x)$ is undefined but $\lim_{h\to0}f(0+h^2)=\lim_{h\to0}f(h^2)=\lim_{h\to0}1=1$, since $h^2\geq0$. You only check for the right limit with the second definition, while the first definition considers both left and right limit.