$\displaystyle y = e^t \cos{t}, x=e^t \sin{t}$
$\displaystyle y''(x+y)^2 = K(xy'-y)$
Well, $\displaystyle y = e^t \cos{t}$, so $\displaystyle y' = e^t \cos{t} - e^t \sin{t}$,
so $\displaystyle y'' = e^t \cos{t} - e^t \sin{t} - e^t \sin{t} - e^t \cos{t} = -2e^t \sin{t}$
$\displaystyle (x + y)^2 = e^{2t} (\cos^2{t} + \sin^2{t}) + 2e^t \sin{t} \cos{t} = e^{2t} + 2e^t \sin{t} \cos{t}$
$\displaystyle xy' = e^t( \cos{t} - \sin{t}) \times e^t \sin{t} = e^{2t} \sin{t} (\cos{t} - \sin{t})$
$\displaystyle y''(x+y)^2 = K(xy'-y)$
$\displaystyle \Rightarrow K = \frac{y''(x + y)^2}{xy' - y}$
$\displaystyle \Rightarrow K = \frac{-2e^t \sin{t} \times (e^{2t} + 2e^t \sin{t} \cos{t})}{e^{2t} \sin{t} (\cos{t} - \sin{t}) - e^t \cos{t}}$
$\displaystyle \Rightarrow K = \frac{-2e^t \sin{t} \times (e^t + 2 \sin{t} \cos{t})}{e^t \sin{t} (\cos{t} - \sin{t}) - \cos{t}}$
Hmmm I can't see this going where we want.....