Discontinuity via two paths

Continuity of $f$. On $y=ax$, $f(x,ax)=\frac{1}{1+a}$. As $x\to0$, this tends to $\frac{1}{1+a}$ for each $a \neq -1$. Hence $f$ is not continuous at the origin.
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Continuity of $g$. On $y=ax$, $g(x,ax)=\frac{ax}{1+a} \rightarrow 0$ as $x\to0$ for any $a \neq -1$ (this does not mean $g$ is continuous!). Along $y=ax+bx^2$ with $a=-1$, $b\neq0$:
$$g(x,-x+bx^2)=\frac{x(-x+bx^2)}{bx^2}=\frac{-x+bx^2}{bx}=\frac{-1}{b}+x \rightarrow \frac{-1}{b}$$
as $x \rightarrow 0$. Limits differ along different paths; $g$ is not continuous at the origin.

Remark. The example of $g$ shows that checking limits along all straight lines through the origin is necessary but *not* sufficient for the existence of a two-variable limit.