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(C) The given differential equation is $(x^3 + 1)dy = x(1 - 3xy)dx$.Rearranging the terms,we get $(x^3 + 1)dy + 3x^2ydx = xdx$.This can be written as

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$2$

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(C) The given differential equation is $(x^3 + 1)dy = x(1 - 3xy)dx$.Rearranging the terms,we get $(x^3 + 1)dy + 3x^2ydx = xdx$.This can be written as $d(y(x^3 + 1)) = xdx$.Integrating both sides,we get $y(x^3 + 1) = \int xdx = \frac{x^2}{2} + C$.Given $f(0) = 0$,substituting $x = 0$ and $y = 0$ gives $0(0 + 1) = 0 + C$,so $C = 0$.Thus,$y(x^3 + 1) = \frac{x^2}{2}$,which implies $f(x) = \frac{x^2}{2(x^3 + 1)}$.Now,we calculate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{x^2}{f(x)} = \mathop {\lim }\limits_{x 	o 0} \frac{x^2}{\frac{x^2}{2(x^3 + 1)}} = \mathop {\lim }\limits_{x 	o 0} 2(x^3 + 1) = 2(0 + 1) = 2$.

If the function $y = f(x)$ satisfies the differential equation $(x^3 + 1)dy = x(1 - 3xy)dx$ and $f(0) = 0$,then $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{f(x)}$ is equal to

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