Irrotational field and scalar potential F=y²i+(3xy+e^(2z))j+2ye^…

Note on the question: with F = y² i + (3xy + e^(2z)) j + 2y e^(2z) k as printed, the k-component of curl F is ∂Q/∂x − ∂P/∂y = 3y − 2y = y ≠ 0, so that field is not conservative. The evident intended field has 2xy in the j-component (a standard textbook problem), so the solution below uses

F = y² i + (2xy + e^(2z)) j + 2y e^(2z) k.

Conservative test

P = y², Q = 2xy + e^(2z), R = 2y e^(2z).

i: ∂R/∂y − ∂Q/∂z = 2e^(2z) − 2e^(2z) = 0
j: ∂P/∂z − ∂R/∂x = 0 − 0 = 0
k: ∂Q/∂x − ∂P/∂y = 2y − 2y = 0

curl F = 0, so F is conservative and F = ∇φ.

Scalar potential

∂φ/∂x = y²          → φ = xy² + f(y,z)
∂φ/∂y = 2xy + f_y = 2xy + e^(2z)   → f_y = e^(2z) → f = y e^(2z) + g(z)
∂φ/∂z = 2y e^(2z) + g'(z) = 2y e^(2z) → g' = 0

Answer: φ = xy² + y e^(2z) + c.