No, a function like this need not have a critical point. However, the only example I could find is a bit complicated and unfortunately not explicit. With a bit of work, one could make this explicit, but it would not be pretty. There should be a simpler example, though...
Let $L$ be the ray from the origin at a $45^\circ$ angle, i.e., defined by $x=y \ge 0$, and let $U$ be the open 1-neighborhood of $L$ in the plane, i.e., all points with distance less than $1$ to $L$. Then there exists a $C^\infty$-diffeomorphism $\phi: U \to U\setminus L$, which is the identity in a neighborhood of $\partial U$. (Note that $U$ and $U\setminus L$ are open sets, and $\phi^{-1}$ obviously does not extend to $L$.) Then we can extend $\phi$ to a $C^\infty$-diffeomorphism $\phi: \mathbb{R}^2 \to \mathbb{R}^2 \setminus L$ by defining it to be the identity outside $U$.
Now let $g(x,y) = x^2-y^2$, which already has the correct limits, but obviously has a critical point at $(0,0)$. Define $f(x,y) = g(\phi(x,y))$. Any horizontal or vertical line either does not intersect $U$ at all, or it intersects it in an interval of length $\le 2$$\le 2\sqrt{2}$, and since $f=g$ outside of $U$, we have that the limits of $f$ and $g$ along horizontal and vertical lines are the same, so $f$ has the desired limits. As a composition of smooth functions, it is smooth, and by the chain rule $f$ does not have any critical points, because $\phi$ does not have any, and the only critical point of $g$ is at $(0,0) \in L$, which is not in the image of $\phi$.