Verify that the given function $y=\sqrt{1+x^{2}}$ is a solution of the differential equation $y^{\prime}=\frac{xy}{1+x^{2}}$.

Given function: $y=\sqrt{1+x^{2}}$
Differentiating both sides with respect to $x$:
$y^{\prime}=\frac{d}{dx}(\sqrt{1+x^{2}})$
Using the chain rule:
$y^{\prime}=\frac{1}{2\sqrt{1+x^{2}}} \cdot \frac{d}{dx}(1+x^{2})$
$y^{\prime}=\frac{1}{2\sqrt{1+x^{2}}} \cdot (2x)$
$y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}$
Now,multiply and divide the right side by $\sqrt{1+x^{2}}$:
$y^{\prime}=\frac{x \cdot \sqrt{1+x^{2}}}{\sqrt{1+x^{2}} \cdot \sqrt{1+x^{2}}}$
$y^{\prime}=\frac{x \cdot y}{1+x^{2}}$
Since the derivative matches the given differential equation,the function $y=\sqrt{1+x^{2}}$ is indeed a solution.