Answer
The maximum is $f(1)=0$ and the minimum is $f(3) =-1/9$
Work Step by Step
Given $$y=\frac{1-x}{x^{2}+3 x}, \quad[1,4]$$
Since
\begin{align*}
\frac{dy}{dx}&= \frac{\frac{d}{dx}\left(1-x\right)\left(x^2+3x\right)-\frac{d}{dx}\left(x^2+3x\right)\left(1-x\right)}{\left(x^2+3x\right)^2}\\
&=\frac{x^2-2x-3}{\left(x^2+3x\right)^2}
\end{align*}
Find the critical points
\begin{align*}
f'(x)&=0\\
\frac{x^2-2x-3}{\left(x^2+3x\right)^2}&=0\\
x^2-2x-3&=0\\
(x+1)(x-3)&=0
\end{align*}
Then $x=-1,\ \ x= 3$. Because $-1\notin [1,4]$, then $f(x)$ has a critical point at $x=3 $, since
\begin{align*}
f(1)&= 0\\
f(3)&= \frac{-1}{9}\\
f(4)&=\frac{-3}{28}
\end{align*}
Then the maximum is $f(1)=0$ and the minimum is $f(3) =-1/9$
