It is given that $2xy\,\dfrac{dy}{dx} = x^{2} - y^{2}$ where $x > 0$, $y < 0$. Using the substitution $u = \dfrac{y}{x}$, show that the differential equation can be transformed to $\dfrac{2u}{1-3u^{2}}\,\dfrac{du}{dx} = \dfrac{1}{x}$. Hence find the general solution of $y$ in terms of $x$.

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

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

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

Substitute into $2xy\frac{dy}{dx} = x^2 - y^2$:
$2x(ux)\left(u + x\frac{du}{dx}\right) = x^2 - (ux)^2$
$2x^2 u\left(u + x\frac{du}{dx}\right) = x^2(1 - u^2)$
Since $x > 0$, we can divide by $x^2$:
$2u^2 + 2ux\frac{du}{dx} = 1 - u^2$
$2ux\frac{du}{dx} = 1 - 3u^2$
$\frac{2u}{1-3u^2}\frac{du}{dx} = \frac{1}{x}$

3. Separate the variables $u$ and $x$ and integrate both sides.

$\int \frac{2u}{1-3u^2} \,du = \int \frac{1}{x} \,dx$
$-\frac{1}{3} \int \frac{-6u}{1-3u^2} \,du = \int \frac{1}{x} \,dx$
$-\frac{1}{3}\ln|1-3u^2| = \ln|x| + c$
Since $x > 0$, $\ln|x| = \ln x$:
$-\frac{1}{3}\ln|1-3u^2| = \ln x + c$

4. Rearrange the equation to express $u^2$ in terms of $x$.

Multiply by $-3$:
$\ln|1-3u^2| = -3\ln x + C'$
$|1-3u^2| = e^{-3\ln x + C'} = e^{C'} x^{-3}$
$1-3u^2 = \frac{C}{x^3} \quad \text{(where } C = \pm e^{C'})$
$3u^2 = 1 - \frac{C}{x^3} = \frac{x^3 - C}{x^3}$

5. Substitute $u = \frac{y}{x}$ back and apply the condition $y < 0$ to find the explicit general solution.

$3\left(\frac{y}{x}\right)^2 = \frac{x^3 - C}{x^3}$
$3y^2 = \frac{x^3 - C}{x}$
$y^2 = \frac{x^3 - C}{3x}$
Since it is given that $x > 0$ and $y < 0$, we must choose the negative square root:
$y = -\sqrt{\frac{x^3 - C}{3x}}$