@NateEldredge As the user has shown that for part A, $f(x,y)$ is discontinuous at $(\frac{1}{\sqrt{2}},\frac{1}{2})$ but in this case, $f_1(x)$ is continuous at $x\in [0,1]$. So, I suppose there is something more to it other than just the replacement of "both" with "either".