Answer
$$y'=\frac{e^{-1/x}}{x^2y}$$
Work Step by Step
$$y^2=2e^{-1/x}$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$2y\times y'=2e^{-1/x}\times\Big(-\frac{1}{x}\Big)'$$
$$2yy'=-2e^{-1/x}\times\frac{(1)'x-1(x)'}{x^2}$$
$$2yy'=-2e^{-1/x}\times\frac{0\times x-1\times1}{x^2}$$
$$2yy'=-2e^{-1/x}\times\frac{-1}{x^2}$$
$$2yy'=\frac{2e^{-1/x}}{x^2}$$
$$yy'=\frac{e^{-1/x}}{x^2}$$
Finally, calculate for $y'$: $$y'=\frac{e^{-1/x}}{x^2y}$$
