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**Hint:**This problem comes under implicit function on differentiation. We need to solve separately and want to differentiate on the function and want to find y and then there will be re arranging and substitute to the equation to compare the coordinates for the solving \[y'(0) = \dfrac{{dy}}{{dx}}\] which first order differentiation and there will be multiple solvable equation and then complete step by step explanation.

**Complete step-by-step answer:**

\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt {5\pi } }}\] \[{{\text{e}}^{\operatorname{Sin} ({x^2} + {y^2})}} = {\text{ tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x - - - - - - - - (1)\]

When \[x = 0\], then the inverse function will be zero and the values of \[x\] will be so, then we get

\[ \Rightarrow {{\text{e}}^{\sin {y^2}}} = {\text{ tan }}\dfrac{{{y^2}}}{4}\]

Taking $\log $ on both sides,

\[ \Rightarrow \sin {y^2} = {\text{ log tan }}\dfrac{{{y^2}}}{4}\]

Hence we get,

\[y = \pm \sqrt x , \pm \sqrt {5x} ,..............\]

Now differentiating equation (1) with respect to x, we get

Let us solve separately Left hand side and Right hand side,

Now differentiate Left hand side

\[ \Rightarrow {{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}}\]

\[ \Rightarrow {{\text{e}}^{\sin \left( {{x^2} + {y^2}} \right)}}\cos \left( {{x^2} + {y^2}} \right)\left( {2x + 2y\dfrac{{dy}}{{dx}}} \right)\]

Now differentiate Right hand side

\[ \Rightarrow {\text{tan }}\dfrac{{{y^2}}}{4} + {\sin ^{ - 1}}x\]

Differentiating we get,

\[ \Rightarrow \dfrac{{2y}}{4}{\sec ^2}\dfrac{{{y^2}}}{4}\dfrac{{dy}}{{dx}} + \dfrac{1}{{\sqrt {1 - {x^2}} }}\]

Now again substitute \[x = 0\] and compare,

\[ \Rightarrow {{\text{e}}^{\sin {y^2}}}\left[ {\cos {y^2}\left( {2y\dfrac{{dy}}{{dx}}} \right)} \right] = \dfrac{y}{2}{\sec ^2}\dfrac{{{y^2}}}{4}\dfrac{{dy}}{{dx}} - - - - (2)\]

Now substitute \[x\] value and \[y\] value in equation (2), we get \[y'(0) = \dfrac{{dy}}{{dx}}\]

Now, at \[x = 0\], \[y = \sqrt \pi \] in equation in (2),

\[ \Rightarrow {{\text{e}}^{\sin {{(\sqrt \pi )}^2}}}[\cos {(\sqrt \pi )^2}2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{\sec ^2}\dfrac{{{{\sqrt \pi }^2}}}{4}\dfrac{{dy}}{{dx}}\]

Now separate \[\dfrac{{dy}}{{dx}}\],

\[ \Rightarrow {{\text{e}}^{\sin \pi }}[\cos \pi ]2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{\sec ^2}\dfrac{\pi }{4}\dfrac{{dy}}{{dx}}\]

We know that \[\sin \pi = 0\] and \[\cos \pi = - 1\],

\[ \Rightarrow {{\text{e}}^0}[ - 1]2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2}{(\sec \dfrac{\pi }{4})^2}\dfrac{{dy}}{{dx}}\]

We know that \[{\sec ^2}\dfrac{\pi }{4}\dfrac{{dy}}{{dx}} = 2\] and \[{e^0} = 1\],

$ \Rightarrow - 2\sqrt \pi \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt \pi }}{2} \times 2\dfrac{{dy}}{{dx}}$

Cancelling the term $2$ in denominator and numerator,

$\Rightarrow$\[ - 2\sqrt \pi \dfrac{{dy}}{{dx}} = \sqrt \pi \dfrac{{dy}}{{dx}}\]

Rearranging the terms we get,

$\Rightarrow$\[ - 2\sqrt \pi \dfrac{{dy}}{{dx}} - \sqrt \pi \dfrac{{dy}}{{dx}} = 0\]

Taking common term same as in both terms,

$\Rightarrow$\[( - 2\sqrt \pi - \sqrt \pi )\dfrac{{dy}}{{dx}} = 0\]

Subtracting the terms we get,

$\Rightarrow$\[ - 3\sqrt \pi \dfrac{{dy}}{{dx}} = 0\]

Hence we get,

$\Rightarrow$\[\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{3\sqrt \pi }}\]

Similarly we can find for when \[x = 0,y = - \sqrt \pi \] we get

$\Rightarrow$\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt \pi }}\]

Similarly we can find for when\[x = 0,y = \sqrt {5\pi } \], we get

$\Rightarrow$\[\dfrac{{dy}}{{dx}} = - \dfrac{1}{{3\sqrt {5\pi } }}\]

Similarly we can find for when \[x = 0,y = - \sqrt {5\pi } \], we get

$\Rightarrow$\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{3\sqrt {5\pi } }}\]

There will be multiple answers and for this question, these are,

$\Rightarrow$$\dfrac{1}{{3\sqrt \pi }}$, $ - \dfrac{1}{{3\sqrt \pi }}$ and $ - \dfrac{1}{{3\sqrt {5\pi } }}$

**$\therefore $ The correct answers are option A) $\dfrac{1}{{3\sqrt \pi }}$, B) $ - \dfrac{1}{{3\sqrt \pi }}$ and D) $ - \dfrac{1}{{3\sqrt {5\pi } }}$**

**Note:**This problem needs attention on differentiation and some trigonometric identities, this kind of problem will be able to solve when the when before differentiation and after differentiation for when find x values and y values for finding the first order differential and then simple basic calculation for that arrange and substitute the value in order to find solution.

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