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Assuming the equation reads
[tex]5y \sin(x^2) = 3x \sin(y^2)[/tex]
Differentiating both sides with respect to x gives
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[5y \sin(x^2)\right] = \dfrac{\mathrm d}{\mathrm dx}\left[3x \sin(y^2)\right] \\\\ 5 \dfrac{\mathrm d}{\mathrm dx}\left[y \sin(x^2)\right] = 3 \dfrac{\mathrm d}{\mathrm dx}\left[x \sin(y^2)\right][/tex]
By the product rule,
[tex]5 \left(\dfrac{\mathrm d}{\mathrm dx}[y] \sin(x^2) + y \dfrac{\mathrm d}{\mathrm dx}\left[\sin(x^2)\right]\right) = 3 \left(\dfrac{\mathrm d}{\mathrm dx}[x] \sin(y^2) + x \dfrac{\mathrm d}{\mathrm dx}\left[\sin(y^2)\right]\right)[/tex]
By the chain rule,
[tex]5 \left(\sin(x^2) \dfrac{\mathrm dy}{\mathrm dx} + y \cos(x^2) \dfrac{\mathrm d}{\mathrm dx}\left[x^2\right]\right) = 3 \left(\sin(y^2) + x \cos(y^2) \dfrac{\mathrm d}{\mathrm dx}\left[y^2\right]\right) \\\\ 5 \left(\sin(x^2) \dfrac{\mathrm dy}{\mathrm dx} + 2xy \cos(x^2)\right) = 3 \left(\sin(y^2) + 2xy \cos(y^2) \dfrac{\mathrm dy}{\mathrm dx}\right)[/tex]
Solve for dy/dx :
[tex]5\sin(x^2) \dfrac{\mathrm dy}{\mathrm dx} + 10 xy \cos(x^2)\right) = 3 \sin(y^2) + 6xy \cos(y^2) \dfrac{\mathrm dy}{\mathrm dx} \\\\ \left(5\sin(x^2)-6xy\cos(y^2)\right) \dfrac{\mathrm dy}{\mathrm dx} = 3 \sin(y^2) - 10 xy \cos(x^2) \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \boxed{\dfrac{3 \sin(y^2) - 10 xy \cos(x^2)}{5\sin(x^2)-6xy\cos(y^2)}}[/tex]