I've written some proofs. Can someone please verify them?
Let $f:\mathbb{R} \backslash\{0\} \longrightarrow \mathbb{R}$ be defined by $\displaystyle{f(x) = \frac{1-x^2}{x}}$
(a) Show that $\displaystyle{\lim_{x \to 0^+} f(x) = \infty}$
Let $(x_n)$ be a sequence in $\displaystyle{\left(0, \frac{1}{2}\right)}$ converging to $0$ and let $M>0$.
Pick $N \in \mathbb{N}$ such that $\forall n > N$, $x_n < \displaystyle{\frac{1}{M}}$.
Then, $\displaystyle{\frac{1}{x_n} > M}$ for every $n > N$.
This implies that $\displaystyle{\frac{1-x_n^2}{x_n}} > \frac{1}{x_n} > M$. (Note that $1-y < 1$ for each $y \in \displaystyle{\left(0, \frac{1}{2}\right)}$)
(b) Show that $\displaystyle{\lim_{x \to \infty} f(x) = - \infty}$
Let $(x_n)$ be a sequence in $(1, \infty)$ converging to $\infty$, and let $M < 0$.
Pick $N \in \mathbb{N}$ such that for all $n > N$, $\displaystyle{x_n > 1- M}$.
Then, $1 - x_n < M$
This implies that $\displaystyle{\frac{1}{x_n} - x_n < 1-x_n<M}$.
That is $\displaystyle{\frac{1-x_n^2}{x_n} < M}$ for all $n > N$
As pointed out in a comment above, one of your inequalities is incorrect in your proof of part (a). Your proof of part (b) is just fine, though it might be worth mentioning why we can assume without loss of generality that all the $x_n>1$.
Let us consider for a moment the inequality $$\frac{1-x^2}{x}>M.\tag{$\star$}$$ If we are assuming that $x>0,$ this is equivalent to $$0>x^2+Mx-1.$$ The discriminant of the quadratic above is readily positive for any real $M,$ so it has two real roots, namely: $$\frac12\left(-M\pm\sqrt{M^2+4}\right)$$ By continuity, one can readily show that $$0>x^2+Mx-1$$ if and only if $$-\frac12\left(M+\sqrt{M^2+4}\right)<x<-\frac12\left(M-\sqrt{M^2+4}\right).$$ Observing/proving that $-\frac12\left(M+\sqrt{M^2+4}\right)<0< -\frac12\left(M-\sqrt{M^2+4}\right)$ for all real $M,$ we find that if $$0<x<-\frac12\left(M-\sqrt{M^2+4}\right),$$ then $(\star)$ holds, whence the result of part (a) follows readily. A similar (but slightly more complicated) argument will also work for part (b), but as I said, your approach is fine.