Yes, "slicing" z= f(x,y) with the plane x= 1 gives . The derivative of that is which is 1+ 4= 5 at y= 2. Of course, z(1, 2)= 2+ 4= 6. Ignoring the x-component, that tangent line would be z= 5(y- 2)+ 6= 5y- 4. Two points on that line are y= 2, z= 6, and y= 3, z= 15-4= 11 or, adding the x= 1 coordinate, (1, 2, 6) and (1, 3, 11). A vector from (1, 2, 6) to (1, 3, 11) which would point in the direction of that line is <1- 1, 3- 2, 11- 6>= <0, 1, 5>.

Notice that we could, as easily, have taken the second point to be at, say, y= 0 which would give z= 5(0)- 4= -4 so our two points would be (1, 2, 6) and (1, 0, -4). A vector from (1, 2, 6) to (1, 0, -4) which would also be pointing along the line (but in opposite direction) would be <1- 1, 0- 2, -4- 6>= <0, -2, -10> and parametric equations for the line would be x= 1, y= 2- 2t, z= 6- 10t. Do you see that these are parametric equations for exactly the same line as x= 1, y= 2+ t, z= 6+ 5t?Indeed, we see that the x-coordinate of this line never changes, which explains why the first coordinate equals 0. This means that the tangent line in question has equation x = 1, y = 2 + t, z = 6 + 5t.

Note that when t= -1/2 these new equations give the point (1, 2+ 1, 6+ 5)= (1, 3, 11) as before. Since both lines contain (1, 2, 6) and (1, 3, 11) they are the same line.