If 4 x+3|y|=5 y, then y as a function of x is

We have,

$4 x+3|y|=5 y$

$\Rightarrow 4 x+3 y=5 y$,  if  $y \geq 0$  and,  $4 x-3 y=5 y$,  if  y < 0

$\Rightarrow y= \begin{cases}2 x, & x \geq 0 \\ \frac{x}{2}, & x<0\end{cases}$

Clearly, y is continuous at x = 0.

We have,

(LHD of y at x = 0) = $\frac{1}{2}$ and, (RHD of y at x = 0) = 2

So, y is not differentiable at x = 0.

Also, $\frac{d y}{d x}= \begin{cases}2, & x>0 \\ \frac{1}{2}, & x<0\end{cases}$