Step 1: Invert the function for a cleaner view.
For positive $x$, look at $\dfrac{1}{f(x)}=\dfrac{1+4x+x^2}{x}=x+\dfrac{1}{x}+4$. Maximizing $f$ means minimizing $\dfrac1f$.
Step 2: Minimize $x+\dfrac1x$.
By AM to GM, for $x>0$, $x+\dfrac1x\ge 2$, with equality at $x=1$.
Step 3: Get the minimum of the reciprocal.
So $\dfrac{1}{f}\ge 2+4=6$, with equality at $x=1$.
Step 4: Translate back to $f$.
Hence the largest value of $f$ for $x>0$ is $\dfrac{1}{6}$.
Step 5: Confirm with a derivative check.
Differentiating, $f'(x)=\dfrac{1-x^2}{(1+4x+x^2)^2}$, which is zero at $x=1$ and changes from positive to negative there, a maximum.
Step 6: Evaluate at $x=1$.
$f(1)=\dfrac{1}{1+4+1}=\dfrac16$.
\[ \boxed{\dfrac{1}{6}} \]