$\underset{x\to 0}{lim}(\frac{1}{x(x+1)}-\frac{\log (1+x)}{x^2})$

Step 1 ：First, we find the derivatives of the functions in the numerator and denominator:

Step 2 ：$\frac{d}{dx}(\log (1+x))=\frac{1}{x+1}$

Step 3 ：$\frac{d}{dx}\left(x^2\right)=2x$

Step 4 ：$\frac{d}{dx}\left(1\right)=0$

Step 5 ：$\frac{d}{dx}(x(x+1))=2x+1$

Step 6 ：Now, we apply L'Hopital's rule and plug in the derivatives:

Step 7 ：$\underset{x\to 0}{lim}(\frac{\frac{1}{x+1}}{2x}-\frac{0}{2x+1})$

Step 8 ：Simplify the expression:

Step 9 ：$\underset{x\to 0}{lim}\left(\frac{1}{2x(x+1)}\right)$

Step 10 ：Finally, find the limit as x approaches 0:

Step 11 ：$\boxed{{\displaystyle \mathrm{\infty }}}$