little help

[mattflint50]

I am going over an old Calculus AB exam. I came across this problem. If x^2-2xy+3y^2=8 then dy/dx equals?

I did it this way but it does not come out.
1.) dy/dx [x^2-2xy+3y^2]=dy/dx [8]
2.) 2x-(2x dy/dx+y)+6y dy/dx=0
3.)2x dy/dx + 6y dy/dx= -2x-y
4.) dy/dx=-2x-y/2x+6y

What am I doin gwrong

 

[Unco]

I am going over an old Calculus AB exam. I came across this problem. If x^2-2xy+3y^2=8 then dy/dx equals?

I did it this way but it does not come out.
1.) dy/dx [x^2-2xy+3y^2]=dy/dx [8]
2.) 2x-(2x dy/dx+y)+6y dy/dx=0

Just one small error when differentiating -2xy.

\(\displaystyle \mbox{ \frac{d}{dx} \left(-2xy\right) = -2x\frac{dy}{dx} -2y}\)

3.)2x dy/dx + 6y dy/dx= -2x-y Be careful with your signs.
4.) dy/dx=-2x-y/2x+6y