The maximum value

Given : $f(x)=\frac{x}{4-x+x^{2}}$

$\Rightarrow f^{\prime}(x)=\frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}}$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow \frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}}=0$

$\Rightarrow 4-x+x^{2}-x(-1+2 x)=0$

$\Rightarrow 4-x+x^{2}+x-2 x^{2}=0$

$\Rightarrow x^{2}=4$

$\Rightarrow x=\pm 2 \notin[-1,1]$

So,

$f(-1)=\frac{-1}{4-(-1)+(-1)^{2}}=\frac{-1}{6}$

$f(1)=\frac{1}{4-1+1^{2}}=\frac{1}{4}$

Hence, the maximum value is $\frac{1}{4}$.

Disclaimer: The question in the book has some error. So, none of the options are matching with the solution. The solution here is according to the question given in the book.