Step 1: Confirm homogeneity by checking degree-0 scaling:
Replacing $x\to tx,\ y\to ty$ leaves $\dfrac{x+2y}{x-y}$ unchanged (both numerator and denominator scale by $t$ and cancel), confirming the equation is genuinely homogeneous and the $v=y/x$ substitution applies.
Step 2: After substitution, isolate $\dfrac{dv}{v^2+v+1}$ on one side symbolically before splitting the numerator:
Following the same algebra, $x(1-v)dv=(v^2+v+1)dx/x$ rearranges to $\dfrac{(1-v)\,dv}{v^2+v+1}=\dfrac{dx}{x}$ — identical separated form.
Step 3: Integrate using the completed-square denominator directly, treating the two pieces of the numerator as one combined rational + one arctan integral:
$v^2+v+1=(v+\tfrac12)^2+\tfrac34$; the $-v$ part of the numerator pairs with the derivative of the denominator (giving a log term with coefficient $-\tfrac12$), while the constant $+1$ part gives a pure arctan term scaled by $\tfrac32\cdot\tfrac{2}{\sqrt3}=\sqrt3$.
Step 4: Substitute $v=y/x$ back and simplify the logarithm of a ratio into a difference of logs, which then cancels against $\ln|x|$ on the right:
This reproduces the same combined implicit relation between $x$ and $y$.
Final Answer:
\[ \boxed{\sqrt3\tan^{-1}\!\left(\dfrac{x+2y}{\sqrt3\,x}\right)-\dfrac12\ln(x^2+xy+y^2)=C} \]