1. y = 1 + y^x
Берем производную по х от обеих частей:
y' = (1 + y^x)'
y' = (e^ln (y^x))'
y' = (e^(x * ln y))'
y' = e^(x * ln y) * (x * ln y)'
y' = y^x * (ln y + x * y'/y)
y' = y^x * ln y + x * y' * y^(x - 1)
y' * (1 - x * y^(x - 1)) = y^x * ln y
y' = y^x * ln y/(1 - x * y^(x - 1))

2. x + y = e^(x - y)
Берем производную по х от обеих частей:
(x + y)' = (e^(x - y))'
1 + y' = e^(x - y) * (x - y)'
1 + y' = e^(x - y) * (1 - y')
1 + y' = e^(x - y) - e^(x - y) * y'
y' + e^(x - y) * y' = e^(x - y) - 1
y' = (e^(x - y) - 1)/(e^(x - y) + 1)

3. y = x + ln y
y' = (x + ln y)'
y' = 1 + y'/y |* y
y * y' = y + y'
y' * (y - 1) = y
y' = y/(y - 1)