1. y=ln(x); dy/dx=1/x; this is the tangent or gradient at point (x,ln(x)). Put in x=e and we get dy/dx=1/e and we get the point (e,1). The equation of the tangent is normally represented as a standard linear equation y=mx+c where m=dy/dx (at a particular point) and c is found by inserting the coords of the point of the tangent. So in the case in question, the point is (e,1) and dy/dx=1/e. y=x/e+c, 1=e/e+c so c=0 and y=x/e is the equation of the tangent line.
2. xy=5 (hyperbola) is differentiated thus: xdy/dx+y=0 (this uses the rule for differentiating a product uv, where u=x and v=y; d/dx(uv)=udv/dx+vdu/dx=xdy/dx+y). So dy/dx=-y/x=-(1/2)/10=-1/20. The equation of the tangent is in the form y=mx+c, where m=-1/20 and c is found by substituting the point coords: 1/2=-10/20+c; c=1/2+1/2=1, so y=1-x/20 or 20y=20-x is the equation of the tangent at (10,1/2).