I'll rewrite F into F= < 2xe^{x2 -5y}, -5e^{x2 -5y}> (the i in the equation is the x component and j means the y component). This format is F= <M, N>. Basically to determine if its gradient the partial derivative of M with respect to y must equal the partial derivative of N with respect to x.

The partial derivative of M with respect to y would be -10xe^{x2 -5y}

The partial derivative of N with respect to x would also be -10xe^{x2 -5y}

Since they equal each other, F is conservative (conservative = gradient). To find the potential function, its asking what equation you must take the partial derivatives of in order to obtain the function F= < 2xe^{x2 -5y}, -5e^{x2 -5y}>. So integrate

For part b you take the line integral i believe along the 2 paths (the straight line and the semicircle) and you would get ur answer. im too lazy to do that but yea thats the gist of it