A landscape designer plans a curved path in a park, with all lengths in metres…

Step 1. The degrees. The top $x^3 + y^3$ has degree $3$. The bottom $3xy^2$ has degree $1 + 2 = 3$. So both have degree $3$ and the equation is homogeneous.

Step 2. Substitute. With $y = vx$: $v + x\dfrac{dv}{dx} = \dfrac{1 + v^3}{3v^2}$. So $x\dfrac{dv}{dx} = \dfrac{1 + v^3 - 3v^3}{3v^2} = \dfrac{1 - 2v^3}{3v^2}$.

Step 3. Separate and integrate. $\dfrac{3v^2\,dv}{1 - 2v^3} = \dfrac{dx}{x}$ gives $-\dfrac{1}{2}\ln|1 - 2v^3| = \ln x + c$. So $x^2(1 - 2v^3) = K$.

Step 4. Return to $x$ and $y$. $x^2 - \dfrac{2y^3}{x} = K$. Multiply by $x$: $x^3 - 2y^3 = Kx$. This is the form the designer wrote, with $C = K$.

Step 5. The constant. At $(1, 1)$: $1 - 2 = C \times 1$, so $C = -1$. Then $2y^3 = x^3 + x$, so $y = \left(\dfrac{x^3 + x}{2}\right)^{1/3}$.

Step 6. Evaluate. At $x = 2\ \text{m}$: $y = \left(\dfrac{10}{2}\right)^{1/3} = 5^{1/3} = 1.7100\ \text{m}$. At $x = 3\ \text{m}$: $y = \left(\dfrac{30}{2}\right)^{1/3} = 15^{1/3} = 2.4662\ \text{m}$.