If $f_{xy}$ and $f_{yx}$ are not required to exist, then obviously we can have a function where $f_{xy}$ is continuous but $f_{yx}$ does not exist, such as $f=|y|$.
Theorem. If $f_{xy}$ and $f_{yx}$ exist everywhere, and $f_{xy}$ is continuous at $(0,0)$, then $f_{yx}$ is continuous at $(0,0)$.
Proof. First, the two mixed partial derivatives are equal at $(0,0)$:
$$\begin{aligned}f_{yx}(0,0)&=\lim_{x\to0}\frac{1}{x}(f_y(x,0)-f_y(0,0))\\&=\lim_{x\to0}\lim_{y\to0}\frac{1}{xy}(f(x,y)-f(x,0)-f(0,y)+f(0,0))\\&=\lim_{x\to0}\lim_{y\to0}\frac1y(f_x(\theta x,y)-f_x(\theta x,0))\\&=\lim_{x\to0}f_{xy}(\theta x,0)\\&=f_{xy}(0,0).\end{aligned}$$
By continuity of $f_{xy}$ at $(0,0)$, given $\varepsilon>0$ there exists $\delta>0$ such that if $|x|+|y|<\delta$ then $$|f_{xy}(x,y)-f_{yx}(0,0)|=|f_{xy}(x,y)-f_{xy}(0,0)|<\frac{\varepsilon}{3}.\qquad(1)$$ Now fix this $(x,y)$. By definition of derivative there exists $x'$ with $|x'|<(\delta-|x|-|y|)/2$ such that $$\left|f_{yx}(x,y)-\frac{1}{x'}(f_y(x+x',y)-f_y(x,y))\right|<\frac{\varepsilon}{3}.\qquad(2)$$ Again, by definition of derivative there exists $y'$ with $|y'|<(\delta-|x|-|y|)/2$ such that $$\left|f_y(x+x',y)-f_y(x,y)-\frac{1}{y'}(f(x+x',y+y')-f(x+x',y)-f(x,y+y')+f(x,y))\right|<\frac{\varepsilon|x'|}{3}.\quad(3)$$ Applying mean value theorem twice, there exist $\eta,\theta\in(0,1)$ such that $$f(x+x',y+y')-f(x+x',y)-f(x,y+y')+f(x,y)=x'y'f(x+\eta x',y+\theta y').\qquad(4)$$ Combining formulas (2–4), we get $$|f_{yx}(x,y)-f_{xy}(x+\eta x',y+\theta y')|<\frac{2\varepsilon}{3}.$$ By formula (1), noticing that $|x+\eta x'|+|y+\theta y'|\le|x|+|x'|+|y|+|y'|<\delta$, we have $$|f_{yx}(x,y)-f_{yx}(0,0)|<\varepsilon.$$ Therefore $f_{yx}$ is continuous at $(0,0)$. QED
