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

$$\begin{gathered} \mathrm{y}=\frac{\mathrm{x}}{1+{\mathrm{x}}^2} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(1+{\mathrm{x}}^2\right)\left(1\right)-\mathrm{x}\left(2\mathrm{x}\right)}{{\left(1+{\mathrm{x}}^2\right)}^2} \\ =\frac{1-{\mathrm{x}}^2}{{\left(1+{\mathrm{x}}^2\right)}^2} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{{\left(1+{\mathrm{x}}^2\right)}^2\left(-2\mathrm{x}\right)-\left(1-{\mathrm{x}}^2\right)\left(2\right)\left(1+{\mathrm{x}}^2\right)2\mathrm{x}}{{\left(1+{\mathrm{x}}^2\right)}^4} \\ \frac{\mathrm{dy}}{\mathrm{dx}} \mathrm{is} \min \mathrm{or} \max \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=0 \\ {\left(1+{\mathrm{x}}^2\right)}^2\left(-2\mathrm{x}\right)-4\mathrm{x}\left(1-{\mathrm{x}}^2\right)\left(1+{\mathrm{x}}^2\right)=0 \\ 2\mathrm{x}\left(-1-{\mathrm{x}}^2-2+2{\mathrm{x}}^2\right)=0 \\ \mathrm{x}=0, \mathrm{x}=\pm \sqrt{3} \\ \therefore \mathrm{x}=0 \mathrm{has} \mathrm{maximum} \\ \mathrm{y}=\frac{\mathrm{x}}{1+{\mathrm{x}}^2}=\frac{0}{1+0}=0 \\ \therefore \mathrm{Point} \left(0, 0\right) \end{gathered}$$