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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