$\lim _{x \rightarrow 0}\left(\frac{1}{x(x+1)}-\frac{\log (1+x)}{x^{2}}\right)$

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

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

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

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

Step 5 ：\(\frac{d}{dx}\left(x(x+1)\right) = 2x+1\)

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

Step 7 ：\(\lim _{x \rightarrow 0}\left(\frac{\frac{1}{x+1}}{2x} - \frac{0}{2x+1}\right)\)

Step 8 ：Simplify the expression:

Step 9 ：\(\lim _{x \rightarrow 0}\left(\frac{1}{2x(x+1)}\right)\)

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

Step 11 ：\(\boxed{\infty}\)