Given $x+|y|=2 y$
$$
\begin{aligned}
& \Rightarrow \quad x+y=2 y \text { or } x-y=2 y \\
& \Rightarrow \quad x=y \text { or } x=3 y
\end{aligned}
$$
This represent a straight line which passes through origin.
Hence, $x+|y|=2 y$ is continuous at $x=0$. Now, we check differentiability at $x=0$
$$
\begin{gathered}
x+|y|=2 y \Rightarrow x+y=2 y, y \geq 0 \\
x-y=2 y, y < 0
\end{gathered}
$$
Thus, $f(x)=\left\{\begin{array}{ll}x, & y < 0 \\ x / 3, & y \geq 0\end{array}\right\}$

$$
\begin{aligned}
& \text { Now, L.H.D. }=\lim _{h \rightarrow 0^{-}} \frac{f(x+h-) f x}{-h}(\quad) \\
& =\lim _{h \rightarrow 0^{-}} \frac{x+h-x}{-h}=-1 \\
& \text { R.H.D }=\lim _{h \rightarrow 0^{+}} \frac{f(x+h-) f x}{h}(\quad) \\
& =\lim _{h \rightarrow 0^{+}} \frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{3}=\frac{1}{3} \\
&
\end{aligned}
$$
Since, L.H.D $\neq$ R.H.D. at $x=0$
$\therefore$ given function is not differentiable at $x=0$