What is the slope of the intersection curve of the superfice $z=xe^{x^2y}$ with the plane $y=\ln2$ when $x=1$ Help with this excercise hehe,, 
ok,,
What is the slope of the  intersection curve of the surface 
$$z=xe^{x^2y}$$
with the plane $$y=\ln2$$ at $x=1$?

If $y=\ln(2)$ then $$z=xe^{x^2\ln2}=x\left[e^{\ln2}\right]^{x^2}=x2^{x^2}.$$
The derivative of this function w.t.r $x$ is
$$2^{x^2}+2x^2\ln(2)2^{x^2}.$$
Substituting $1$ for $x$ we get
$$2+4\ln(2),$$
the slope of the red line in the figure below

