Answer
The maximum is $f(0)=f(3)=0$ and the minimum is $f(1) =-1$
Work Step by Step
Given $$y=x-\frac{4 x}{x+1}, \quad[0,3]$$
Since
\begin{align*}
\frac{dy}{dx}&=1-\frac{4}{\left(x+1\right)^2}\\
&=\frac{x^2+2x-3}{\left(x+1\right)^2}
\end{align*}
Find the critical points
\begin{align*}
f'(x)&=0\\
\frac{x^2+2x-3}{\left(x+1\right)^2}&=0\\
x^2+2x-3&=0\\
(x-1)(x+3)&=0
\end{align*}
Then $x=1,\ \ x= -3$. Because $-3\notin [0,3]$, then $f(x)$ has a critical point at $x=1 $, since
\begin{align*}
f(0)&= 0\\
f(3)&=0 \\
f(1)&= -1
\end{align*}
Then the maximum is $f(0)=f(3)=0$ and the minimum is $f(1) =-1$
