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**Hint:**Derivatives are defined as the varying rate of a function with respect to an independent variable. To differentiate the right hand side of the equation we use the product rule. That is if we have \[y = uv\] then its differentiation with respect to ‘x’ is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\]. We solve this using implicit differentiation. We know that the differentiation of \[{e^x}\] is \[{e^x}\] with respect to ‘x’.

**Complete step by step solution:**

Given, \[{e^y} = x{y^2}\]

Now differentiate both sides with respect to ‘x’.

\[\dfrac{d}{{dx}}({e^y}) = \dfrac{d}{{dx}}(x{y^2})\]

Now applying the product rule\[\dfrac{{d(uv)}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times

\dfrac{{du}}{{dx}}\]

Where \[u = x\] and \[u = {y^2}\]. Then we have

\[\dfrac{d}{{dx}}({e^y}) = x \times \dfrac{{d({y^2})}}{{dx}} + {y^2} \times \dfrac{{d(x)}}{{dx}}\]

Applying the differentiation

\[{e^y}\dfrac{{dy}}{{dx}} = x \times 2y\dfrac{{dy}}{{dx}} + {y^2} \times 1\]

\[{e^y}\dfrac{{dy}}{{dx}} = 2xy\dfrac{{dy}}{{dx}} + {y^2}\]

Grouping \[\dfrac{{dy}}{{dx}}\] in on the left hand side of the equation,

\[{e^y}\dfrac{{dy}}{{dx}} - 2xy\dfrac{{dy}}{{dx}} = {y^2}\]

Taking \[\dfrac{{dy}}{{dx}}\] common, we have:

\[\dfrac{{dy}}{{dx}}\left( {{e^y} - 2xy} \right) = {y^2}\]

Dividing the whole equation by \[\left( {{e^y} - 2xy} \right)\],

**\[\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{\left( {{e^y} - 2xy} \right)}}\]. This is the required answer.**

**Note:**We know the differentiation of \[{x^n}\] with respect to ‘x’ is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\]. We know the differentiation of \[{y^n}\] with respect to ‘x’ is

\[\dfrac{{d({y^n})}}{{dx}} = n.{y^{n - 1}}\dfrac{{dy}}{{dx}}\]. We also have different rules in the differentiation. Those are

\[ \bullet \] Linear combination rules: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as \[h'(x) = af'(x) + bg'(x)\]

\[ \bullet \] Product rule: When a derivative of a product of two function is to be found, then we use product rule that is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times

\dfrac{{du}}{{dx}}\].

\[ \bullet \]Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is \[fog'({x_0}) = [(f'og)({x_0})]g'({x_0})\]. We use these rules depending on the given problem.

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