Explanation
\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\,\,\,\,\,\,\,\left( {given} \right)
\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{{dx}}{{{x^2}}}} }
\Rightarrow \log y = - \dfrac{1}{x} + c
\Rightarrow y = {e^{ - \frac{1}{x} + c}}
\Rightarrow y = {e^{ - \frac{1}{x}}},c
since this curve is passing through point \left( {1,3} \right)
so, 3 = {e^{ - \dfrac{1}{t}}}.c
\Rightarrow c = 3e
therefore , eqn of the curve is
y = {e^{ - \dfrac{1}{x}}}.3e
\Rightarrow y = 3{e^{ - \dfrac{1}{x} + 1}}
\Rightarrow y = 3{e^{1 - \dfrac{1}{x}}}
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