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#