Can someone help me with this other limit?

You can use the Taylor expansion to evaluate. In fact, since $$ sin x=x-frac16x^3+O(x^5),tanh x=x-frac13x^3+O(x^5),e^x-1=x+frac12x^2+frac16x^3+O(x^4), ln(1+x)=x-frac12x^2+frac13x^3+O(x^4) $$ then you have $$ lim_{x to 0} frac{(sin x - tanh x)^2}{(e^{x}-1-ln{(1+x)})^3}=lim_{x to 0} frac{bigg[(x-frac16x^3)-(x-frac13x^3)+O(x^5)bigg]^2}{bigg[(x+frac12x^2+frac16x^3)-(x-frac12x^2+frac13x^3)+O(x^4)bigg]^3}=frac{frac{1}{6^2}}{ 1}= frac{1}{36}. $$