A family of curves is defined by the differential equation $xy\,\frac{dy}{dx} = x^{2} + y^{2}.$

1. Differentiate the given substitution $y = ux$ with respect to $x$.

$\frac{dy}{dx} = u + x\frac{du}{dx}$

2. Substitute $y$ and $\frac{dy}{dx}$ into the given differential equation and expand.

$x(ux)\left(u + x\frac{du}{dx}\right) = x^2 + (ux)^2$
$u^2x^2 + ux^3\frac{du}{dx} = x^2 + u^2x^2$

3. Simplify to isolate the derivatives and separate the variables.

$ux^3\frac{du}{dx} = x^2$
$u\frac{du}{dx} = \frac{1}{x}$

4. Integrate both sides with respect to $x$.

$\int u \, du = \int \frac{1}{x} \, dx$
$\frac{1}{2}u^2 = \ln |x| + C$

5. Substitute $u = \frac{y}{x}$ back into the equation and express it in the form $y^2 = f(x)$.

$\frac{1}{2}\left(\frac{y}{x}\right)^2 = \ln |x| + C$
$\frac{y^2}{x^2} = 2\ln |x| + 2C$
$y^2 = x^2(2\ln |x| + A) \quad \text{where } A = 2C$

1. Derive the general solution for the family of curves (from the previous part).

Let $y = ux$, then $\frac{dy}{dx} = u + x\frac{du}{dx}$.
Substitute into $xy\frac{dy}{dx} = x^2 + y^2$:
$x(ux)\left(u + x\frac{du}{dx}\right) = x^2 + (ux)^2$
$u^2x^2 + ux^3\frac{du}{dx} = x^2(1 + u^2)$
$ux\frac{du}{dx} = 1 \implies u\frac{du}{dx} = \frac{1}{x}$
Integrating both sides:
$\int u \, du = \int \frac{1}{x} \, dx$
$\frac{1}{2}u^2 = \ln|x| + c$
$\frac{1}{2}\left(\frac{y}{x}\right)^2 = \ln|x| + c \implies y^2 = 2x^2\ln|x| + Cx^2$

2. Use the given point to find the particular solution.

Substitute $x = 1, y = 2$ into the general solution:
$2^2 = 2(1)^2\ln|1| + C(1)^2$
$4 = 0 + C \implies C = 4$
The particular solution curve is:
$y^2 = 2x^2\ln|x| + 4x^2$

3. Find the values of x when y = 3.

Substitute $y = 3$ into the particular solution:
$3^2 = 2x^2\ln|x| + 4x^2$
$2x^2\ln|x| + 4x^2 - 9 = 0$
Using a graphing calculator to solve for $x$:
$x \approx 1.39 \text{ or } x \approx -1.39 \text{ (to 3 s.f.)}$