The point on the curve   $y=\frac{x}{1+x^2}$  where the tangent to the curve has the greatest slope is

$$\begin{gathered} \frac{dy}{dx}=\frac{\left(1+x^2\right)-2x^2}{{\left(1+x^2\right)}^2} \\ =\frac{1}{1+x^2}-\frac{2x^2}{{\left(1+x^2\right)}^2}=\frac{(1-x{)}^2}{{\left(1+x^2\right)}^2} \\ \frac{d^{2}y}{d{x}^2}=\frac{{\left(1+x^2\right)}^2[2x-4x]-\left[(1+x{)}^2-2x^2\right]\left[2\left(1+x^2\cdot 2x\right]\right.}{{\left(1+x^2\right)}^4} \\ =\frac{{\left(1+x^2\right)}^2[-2x]+\left[2x^2\left(1+x^2\right)\right]\left[4x\left(1+x^2\right)\right]}{{\left(1+x^2\right)}^4}<0 \\ \frac{dy}{dx}=0⟹\left(1-x^2\right)=0⟹x=1 \\ \mathrm{y}=\frac{1}{1+1}=\frac{1}{2} \end{gathered}$$

$\text{ Therefore, point is }(\mathrm{x},\mathrm{y})=\left(1,\frac{1}{2}\right)$