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.