Consider the function [math]f\left(x,y\right)=x^2y+y^2[/math] and the point [math]P\left(1,-1\right)[/math]. Note that [math]f_x=2xy[/math] and [math]f_y=x^2+2y[/math], so that [math]f_x\left(P\right)=-2[/math] and [math]f_y\left(P\right)=-1[/math]. These are the slopes of the tangent lines to the cross sections gotten by fixing y and x, respectively. In the graphs below, I show you the graph, the point in question (1,-1,0), the cross section in black, and the tangent line in light blue.