student grade 12 calculus

write an equation of the line tangent to the graph of

e^{y} + ln(xy) = 1 + e at (e,1)

please help cant figure out

Hi,
I first checked to see if (e,1) satisfies the equation using substitution

e^{1} + ln(e) = e + 1
Hence (e,1) is on the graph. Thus the tangent line is the line through (e,1) with slope given by the derivative of the expression at the point (e,1). You can find the derivative using implicit differentiation.
Differentiating both sides of e^{y} + ln(xy) = 1 + e I get

Can you complete the problem now?

Harley