Before beginning, you should know that derivative of a function of
the form g(x)/f(x) is given by (f(x)*g'(x)-g(x)*f'(x))/(f(x))^2
i.e. if a(x)=g(x)/f(x) then,
color(red)(d/dxa(x) = (f(x)*d/dxg(x)-g(x)*d/dxf(x))/(f(x))^2)
Coming back to the question,
At x=-1; f(x)=(1-2-1)/(4+3)=-2/7
therefore we have to find equation of normal at (-1, -2/7)
slope m_1 of f(x) at any general point (x, y) is given by f'(x)
i.e. m_1 = f'(x)
= ((4-3x)*d/dx(x^2+2x-1)-(x^2+2x-1)*d/dx(4-3x))/(4-3x)^2
=((4-3x)(2x+2)-(x^2+2x-1)(-3))/(4-3x)^2
therefore slope of f(x) at x=-1 is given by m_1]_(x=-1)
m_1]_(x=-1)=((4+3)(-cancel(2)+cancel(2))-(1-2-1)(-3))/(4+3)^2
implies m_1]_(x=-1)= (3*(-2))/49 = -6/49
Let the slope of normal at the given point be m_2.
When two curves are normal/perpendicular to each other at some point then the product of their slopes at that point equals -1.
therefore at (-1, -2/7); m_1*m_2=-1 implies m_2=-1/m_1 = 49/6
Hence equation of normal at given point can be written in slope point form as
y-(-2/7)=m_2(x-(-1))
y+2/7=49/6*(x+1) = 49/6x+49/6
y-49/6x-331/42=0 is the required equation of normal.