Specifying the point #(1,5)# makes C become a defined constant:
#C = (1 - 5^2)/(3(1)(e^(5^2))) = (-24)/(3e^25) = -8e^(-25)#
However, this has no bearing on the slope of the tangent line, because it will be 0, when we differentiate:
#(d(-8e^(-25)))/dx = 0#
Use the Quotient Rule to differentiate the left side:
#f(x,y) = g(x,y)/(h(x,y))#, then #f'(x,y) = {g'(x,y)h(x,y) -g(x,y)h'(x,y)}/{h(x,y)}^2#
Let #g(x,y) = x - y^2#, then #g'(x,y) = 1 - 2ydy/dx#
Let #h(x,y) = 3xe^(y^2)#, then #h'(x,y) = 3e^(y^2) + 6xye^(y^2)dy/dx#
This is equal to 0, therefore, we can rid ourselves of the denominator by multiplying both sides by it.
Substitute the above into the numerator and set it equal to 0:
#(1 - 2ydy/dx)(3xe^(y^2)) - (x - y^2)(3e^(y^2) + 6xye^(y^2)dy/dx) = 0#
#(1 - 2ydy/dx)(3xe^(y^2)) + (y^2 - x)(3e^(y^2) + 6xye^(y^2)dy/dx) = 0#
#3xe^(y^2) - 6xye^(y^2)dy/dx + 3(y^2 - x)e^(y^2) + 6xy(y^2 - x)e^(y^2)dy/dx) = 0#
#3y^2e^(y^2) - 6xye^(y^2)dy/dx + 6xy(y^2 - x)e^(y^2)dy/dx = 0#
#3y^2e^(y^2) + 6xy(y^2 - x - 1)e^(y^2)dy/dx = 0#
#6xy(y^2 - x - 1)e^(y^2)dy/dx = 3y^2e^(y^2)#
#dy/dx = (3y^2e^(y^2))/(6xy(y^2 - x - 1)e^(y^2))#
#dy/dx = (y)/(2x(y^2 - x - 1))#
The slope, m, is the above evaluated at the point #(1,5)#:
#m = (5)/(2(1)(5^2 - 1 - 1))#
#m = 5/(2(23))#
#m = 5/46#
