Find $ \lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) $ with Taylor

You just have to write the Taylor expansion of $\ln \left( 1 + \frac{1}{x} \right)$ when $x$ tends to $+\infty$ : $$x - x^2 \ln \left( 1 + \frac{1}{x} \right) = x - x^2 \left( \frac{1}{x} - \frac{1}{2x^2} + o\left( \frac{1}{x^2} \right) \right) = \frac{1}{2} + o(1)$$

So the limit is equal to $\frac{1}{2}$.

Proceeding with your substitution, since $\log \left( 1+y\right) =y-\frac{1}{ 2}y^{2}+O\left( y^{3}\right) $, we have:

\begin{eqnarray*} \frac{1}{y}-\frac{1}{y^{2}}\log \left( 1+y\right) &=&\frac{1}{y}-\frac{1}{ y^{2}}\left( y-\frac{1}{2}y^{2}+O\left( y^{3}\right) \right) &=&\frac{1}{2}+O\left( y\right) \overset{y\rightarrow 0}{\longrightarrow } \frac{1}{2}. \end{eqnarray*}