Use implicit differentiation to find the slope of the tangent line to the curve

Use implicit differentiation to find the slope of the tangent line to the curve:

(y/(x-7y))=x^(9)-9 at the point (1, -8/-55).

m=______?

y/(x - 7y) = x^9 - 9

Differentiate both sides with respect to x:

d[ y/(x - 7y) ]/dx = d[ x^9 - 9 ]/dx

The Quotient Rule:

d[ m/n ]/dx = [ (n)(dm/dx) - (m)(dn/dx) ] / n^2

d[ y/(x - 7y) ]/dx = [ (x - 7y)(dy/dx) - (y)(d[x]/dx - 7d[y]/dx) ] / (x - 7y)^2
d[ y/(x - 7y) ]/dx = [ (x - 7y)(y'y) - (y)(1 - 7y'y) ] / (x - 7y)^2
d[ y/(x - 7y) ]/dx = y[ (x - 7y)y' - (1 - 7y'y) ] / (x - 7y)^2

y[ (x - 7y)y' - (1 - 7y'y) ] / (x - 7y)^2 = 9x^8

Coordinates:

{ x = 1
{ y = 8/55 ... (The exact value of the original is unclear.)

Substitute:

(8/55)[ (1 - 7(8/55))y' - (1 - 7y'(8/55)) ] / (1 - 7(8/55))^2 = 9^8

Simplify:

(8/55)[ (55/55 - 56/55)y' - (55/55 - y'56/55) ] / (55/55 - 56/55)^2 = 9^8
(8/55)(1/55)[ (55 - 56)y' - (55 - 56y') ] / [ (55 - 56)/(55) ]^2 = 9^8
8[ (-1)y' - (55 - 56y') ] / [ (-1)]^2 = 9^8
-8[ y' + 55 - 56y' ] = 9^8
-8[ 55 - 55y' ] = 9^8
-8(55)[ 1 - y' ] = 9^8
1 - y' = -(9^8)/(8*55)
y' = (9^8)/440 + 1

m = y' = (9^8)/440 + 1