For those who don't want to stand on their heads, the problem says,
"The normal to the graph of y= \frac{2x}{3x+ 1} at (-1, 1) meets the curve again at point B. Find
(i) the equation of the normal.
(ii) the coordinates of B."
If the problem is not important enough to you to simply type that in, why would it be important enough for us to try to read and solve it?
The derivative of \frac{2x}{3x+ 1}, at any point, (x, y), is \frac{2(3x+ 1)- 3(2x)}{(3x+ 1)^2}= \frac{2}{(3x+ 1)^2}. At (-1, 1) that is \frac{1}{2} so the slope of the normal line is -2. That part you have right. However, if you put x= -1 in the equation you give, y= -2x+ \frac{3}{2}, you get y= -2(-1)+ \frac{3}{2}= \frac{7}{2}, NOT 1. You seem to have the idea that you find the equation of the normal by finding the equation of the tangent line, then replacing the slope, m, with -1/m. That is NOT correct. The slope of the normal line is -1/m but the constant term, "b" in "y= mx+ b", also has to be changed to give the correct point.
With y= -2x+ b, to get y= 1 when x= -1, you must have 1= 2+ b so b= -1. The equation of the normal line is y= -2x- 1.
That line will cross y= \frac{2x}{3x+ 1} when -2x+ 1= \frac{2x}{3x+ 1}. If you multiply both sides of that equation by the denominator on the right, you get a quadratic equation. The line clearly crosses the graph of y= \frac{2x}{3x+ 1} in two places with the x value of those two points the two roots of the quadratic equation. One solution is the given point, (1, -1), the other is B.
