After the most obvious simplifications you have
$\begin{align*} h''(x)&=\frac{\sqrt{x^2+1} - x\left(\frac{1}{2}(x^2+1)^{-\frac12}(2x)\right)}{(\sqrt{x^2+1})^2}\\ &=\frac{\sqrt{x^2+1}-x^2(x^2+1)^{-\frac12}}{x^2+1}\\ &=\frac{\sqrt{x^2+1}-\dfrac{x^2}{\sqrt{x^2+1}}}{x^2+1}\;. \end{align*}$
You can now either multiply the fraction directly by $\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}$ or continue simplifying the numerator. If you do the latter, you get
$\begin{align*} \frac{\sqrt{x^2+1}-\dfrac{x^2}{\sqrt{x^2+1}}}{x^2+1}&=\frac{\dfrac{x^2+1-x^2}{\sqrt{x^2+1}}}{x^2+1}\\ &=\frac{\dfrac1{\sqrt{x^2+1}}}{x^2+1}\\ &=\frac1{(x^2+1)\sqrt{x^2+1}}\\ &=\frac1{(x^2+1)^{3/2}}\\ &=(x^2+1)^{-\frac32}\;; \end{align*}$
if you multiply by $\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}$ directly, you end up with the same result after similar calculations.
Note that you could also have used the product rule to calculate $h''(x)$, leaving $h'(x)$ in the form $x(x^2+1)^{-1/2}$. Then
$\begin{align*} h''(x)&=x\left(-\frac12(x^2+1)^{-3/2}(2x)\right)+(x^2+1)^{-1/2}\\ &=-\frac{x^2}{(x^2+1)^{3/2}}+\frac1{(x^2+1)^{1/2}}\\ &=-\frac{x^2}{(x^2+1)^{3/2}}+\frac{x^2+1}{(x^2+1)^{3/2}}\\ &=\frac1{(x^2+1)^{3/2}}\;, \end{align*}$
with perhaps a little less work.