What is dy/dx equal to, given that y = y₀ when x = 1? (Given (e^y)^x - y = 0)

From dy/dx = y²/(1-xy) and at x = 1, y = y₀: dy/dx = y₀²/(1-y₀). From xy = ln y at x=1: y₀ = ln y₀, so y₀ = ln y₀ means e^y₀ = y₀ (i.e., y₀ = e^y₀ if interpreting differently, but from xy=ln y at x=1: y₀ = ln y₀, hence y₀ = ln y₀ giving e^(y₀) = y₀ wait: ln y₀ = y₀ means y₀ = e^y₀ implies... actually e^(y₀) = y₀^? Let me redo: xy = ln y, at x=1, y=y₀: 1·y₀ = ln y₀, so ln y₀ = y₀, i.e., y₀ = e^(y₀)... that gives no real solution. Re-examine: (e^y)^x = e^(xy), so e^(xy) = y, meaning xy = ln y. At x=1, y₀ = ln y₀ has no real solution. Treating algebraically: dy/dx = y²/(1-xy) at x=1, y=y₀: = y₀²/(1-y₀). Using xy = ln y, so 1-y₀ = 1 - ln y₀; with e^y₀ relations the answer simplifies to y₀·e^y₀/(1-e^y₀) per the option matching.