(a) 2(x-2y)(1-2dy/dx)=dy/dx.
2x-4xdy/dx-4y+8ydy/dx=dy/dx,
(dy/dx)(-4x+8y-1)=4y-2x,
dy/dx=(4y-2x)/(8y-4x-1).
(b) This can be written y2-x2=5xy.
2ydy/dx-2x=5xdy/dx+5y,
(2y-5x)dy/dx=2x+5y, dy/dx=(2x+5y)/(2y-5x).
When x=0, y=0, so (0,0) is a point on the graph of this equation. At this point dy/dx is undefined.
y2-5xy-x2=0, so using the quadratic formula:
y=(5x±√(25x2+4x2)/2=(5x±x√29)/2=(5±√29)x, which represents two linear graphs:
y=(5+√29)x/2 and y=(5-√29)x/2. The slopes (gradients) of these two components are perpendicular because (5+√29)(5-√29)/4=-4/4=-1. Therefore there are two slopes: 5.1926 and -0.1926 at right angles. The graphed equation looks like the x and y axes have been tilted clockwise through an angle of about 10.9°. The gradient is therefore never 0 (horizontal) or infinity (vertical) because the inclination of the lines is constant at 10.9° clockwise to the axes.