Let us consider a curve, $$y = f(x)$$ passing through the point $$(-2, 2)$$ and the slope of the tangent to the curve at any point $$(x, f(x))$$ is given by $$f(x) + xf'(x) = x^2$$. Then

We are told that the curve satisfies the differential relation $$f(x)+x\,f'(x)=x^{2}.$$

First, we rewrite it so that the coefficient of $$f'(x)$$ is 1. Dividing every term by $$x$$ (we assume $$x\neq 0$$ while differentiating), we obtain

$$f'(x)+\frac{1}{x}\,f(x)=x.$$

This is a linear first-order differential equation of the standard form

$$y'+P(x)\,y=Q(x),$$

where here $$y=f(x),\; P(x)=\dfrac{1}{x},\; Q(x)=x.$$

For such an equation, the integrating factor is defined by

$$\mu(x)=e^{\int P(x)\,dx}.$$

We calculate the integral of $$P(x):$$

$$\int \frac{1}{x}\,dx=\ln|x|.$$

Hence

$$\mu(x)=e^{\ln|x|}=|x|.$$

For simplicity we take $$\mu(x)=x$$ (the sign has no effect on the integration that follows). Multiplying the entire differential equation by this integrating factor, we get

$$x\,f'(x)+f(x)=x^{2}.$$

Now observe that the left‐hand side is precisely the derivative of the product $$x\,f(x)$$ with respect to $$x$$, since

$$\frac{d}{dx}\bigl[x\,f(x)\bigr]=x\,f'(x)+f(x).$$

Therefore the differential equation reduces to the very simple form

$$\frac{d}{dx}\,[x\,f(x)]=x^{2}.$$

We integrate both sides with respect to $$x$$:

$$x\,f(x)=\int x^{2}\,dx=\frac{x^{3}}{3}+C,$$

where $$C$$ is the constant of integration. Solving for $$f(x)$$ gives

$$f(x)=\frac{x^{2}}{3}+\frac{C}{x}.$$

The curve passes through the point $$(-2,\,2)$$, so we substitute $$x=-2$$ and $$f(x)=2$$ to find $$C$$:

$$2=\frac{(-2)^{2}}{3}+\frac{C}{-2}=\frac{4}{3}-\frac{C}{2}.$$

Clearing denominators by multiplying every term by $$6$$:

$$12=8-3C.$$

Thus

$$3C=-4\;\Longrightarrow\;C=-\frac{4}{3}.$$

We now insert this value back into the expression for $$f(x)$$:

$$f(x)=\frac{x^{2}}{3}-\frac{4}{3x}.$$

To eliminate the denominators, multiply both sides by $$3$$:

$$3\,f(x)=x^{2}-\frac{4}{x}.$$

Next multiply every term by $$x$$ to remove the remaining fraction:

$$3x\,f(x)=x^{3}-4.$$

Finally, bring all terms to one side to obtain a purely algebraic relation between $$x$$ and $$f(x)$$:

$$x^{3}-3x\,f(x)-4=0.$$

This matches the expression given in Option A. None of the other options coincide with the derived relation.

Hence, the correct answer is Option A.