What is y'= at point (1,-1) when the equation is 3y^2-4x^2+1=0

Implicit Differentiation! Please Help Me!

note that : d(u^n)/dx=nu^(n-1)*(du/dx)=nu^(n-1) u'
d(3y^2)=2*3*y *y' = 6yy'
d(-4x^2)=-8x
d(1)=0
---->6yy'-8x=0----> y'=8x/(6y)
----> y'(at (1,-1)) = (8*1)/(6*-1)=8/-6=-4/3

Differentiate first
3y^2-4x^2+1=0.
6y (y') - 8x = 0
=> y'= (4/3)(x/y).
At (x, y) = (1, -1),
y'= (4/3)(x/y)
= (4/3)(1/-1)
= - 4/3
= m .
It is a tangent (straight line) with the equation
y = - (4/3)x + c
gives c = 1/3
=> 4x + 3y - 1= 0 .