已知 $2xy\dfrac{dy}{dx} = x^{2} - y^{2}$，其中 $x>0$，$y<0$。通过换元 $u = \dfrac{y}{x}$，证明该微分方程可以转化为 $\dfrac{2u}{1-3u^{2}}\dfrac{du}{dx} = \dfrac{1}{x}.$ 由此求出 $y$ 关于 $x$ 的通解。

1. Differentiate the given substitution $u = \frac{y}{x}$ with respect to $x$.

Rearranging $u = \frac{y}{x}$, we have $y = ux$.
Differentiating with respect to $x$ using the product rule:
$\frac{dy}{dx} = u + x\frac{du}{dx}$

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

Substitute $y$ and $\frac{dy}{dx}$ 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 $x > 0$, divide both sides by $x^{2}$:
$2u^{2} + 2ux\frac{du}{dx} = 1 - u^{2}$
$2ux\frac{du}{dx} = 1 - 3u^{2}$
Dividing by $x(1-3u^{2})$ gives:
$\frac{2u}{1-3u^{2}}\frac{du}{dx} = \frac{1}{x} \quad \text{(Shown)}$

3. Separate the variables to set up the integrals.

Separating the variables $u$ and $x$:
$\int \frac{2u}{1-3u^{2}} \, du = \int \frac{1}{x} \, dx$

4. Integrate both sides.

For the left hand side, rewrite the numerator to match the derivative of the denominator:
$-\frac{1}{3} \int \frac{-6u}{1-3u^{2}} \, du = \int \frac{1}{x} \, dx$
$-\frac{1}{3}\ln|1-3u^{2}| = \ln x + C$
(Note: $\ln|x|$ simplifies to $\ln x$ since we are given $x > 0$)

5. Eliminate the logarithms to express $u$ algebraically.

Multiply by $-3$ and express as a single exponential equation:
$\ln|1-3u^{2}| = -3\ln x - 3C = \ln(x^{-3}) + C_{1}$
$|1-3u^{2}| = e^{C_{1}}x^{-3}$
Removing the modulus by introducing a new arbitrary constant $A = \pm e^{C_{1}}$:
$1-3u^{2} = \frac{A}{x^{3}}$

6. Substitute back $u = \frac{y}{x}$ and solve explicitly for $y$.

Substitute $u = \frac{y}{x}$ back into the equation:
$3\left(\frac{y}{x}\right)^{2} = 1 - \frac{A}{x^{3}}$
$\frac{3y^{2}}{x^{2}} = \frac{x^{3}-A}{x^{3}}$
$y^{2} = \frac{x^{3}-A}{3x}$
Since we are given $y < 0$, we must take the negative square root:
$y = -\sqrt{\frac{x^{3}-A}{3x}}$