The coordinates of the point $\mathrm{P}(\mathrm{x},\mathrm{y})\text{ of }\mathrm{y}={\mathrm{e}}^{-\left|\mathrm{x}\right|}$ so that the area formed by the coordinate axes and the tangent at P is greatest are

Graph of  $\mathrm{y}={\mathrm{e}}^{-\left|\mathrm{x}\right|}$ is symmetrical about y-axis, so we consider $\mathrm{x}\ge 0,$ then 

$\mathrm{y}={\mathrm{e}}^{-\left|\mathrm{x}\right|}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-{\mathrm{e}}^{\mathrm{x}}$

Tangent at P is $\mathrm{Y}-\mathrm{y}=-{\mathrm{e}}^{-\mathrm{x}}(\mathrm{X}-\mathrm{x})$

Its x-intercept$=\mathrm{x}+{\mathrm{ye}}^{\mathrm{x}}$ and y-intercept$=\mathrm{y}+{\mathrm{xe}}^{-\mathrm{x}}$

So, area $\mathrm{A}=\frac{1}{2}\left(\mathrm{x}+{\mathrm{ye}}^{\mathrm{x}}\right)\left(\mathrm{y}+{\mathrm{xe}}^{-\mathrm{x}}\right)$

$\begin{array}{l}=\frac{1}{2}(\mathrm{x}+1)(1+\mathrm{x}){\mathrm{e}}^{-\mathrm{x}}\left[\therefore \mathrm{y}={\mathrm{e}}^{-\mathrm{x}}\right]\\ =\frac{1}{2}(\mathrm{x}+1{)}^2{\mathrm{e}}^{-\mathrm{x}}\\ \frac{\mathrm{dA}}{\mathrm{dx}}=(\mathrm{x}+1){\mathrm{e}}^{-\mathrm{x}}-\frac{1}{2}(\mathrm{x}+1{)}^2{\mathrm{e}}^{-\mathrm{x}}=\frac{1}{2}(\mathrm{x}+1\left)\right(1-\mathrm{x}){\mathrm{e}}^{-\mathrm{x}}\end{array}$

Where A is maximum id x=1

So, P is $\left(1,{\mathrm{e}}^{-1}\right)$ Due to symmetry, there is another point $\left(-1,{\mathrm{e}}^{-1}\right)$