Find the tangent to this curve

A curve is given by
3x^2 +3y^2 + 2xy − 2x − 2y = 0

what is the tangent to this curve at (0,0)

Is it y=-x?

Thanks in advance for answering

Let's find out

3x^2 + 3y^2 + 2xy - 2x - 2y = 0
6x * dx + 6y * dy + 2x * dy + 2y * dx - 2 * dx - 2 * dy = 0

x = 0
y = 0

6 * 0 * dx + 6 * 0 * dy + 2 * 0 * dy + 2 * 0 * dx - 2 * dx - 2 * dy = 0
0 * dx + 0 * dy + 0 * dy + 0 * dx - 2 * dx - 2 * dy = 0
-2dx - 2dy = 0
dx + dy = 0
dy = -dx
dy/dx = -1

So the slope is -1 and passes through (0 , 0)

y = -x + b
0 = 0 + b
0 = b

y = -x

You got it right.

Use implicit differentiation.
d/dx(3x^2 + 3y^2 + 2xy - 2x - 2y) = d/dx(0)
6x + 6y(dy/dx) + 2x(dy/dx) + 2y - 2 - 2(dy/dx) = 0
6y(dy/dx) + 2x(dy/dx) - 2(dy/dx) = 2 - 6x - 2y
dy/dx(6y + 2x - 2) = 2 - 6x - 2y
dy/dx = (2 - 6x - 2y)/(6y + 2x - 2)
dy/dx(0,0) = (2 - 6(0) - 2(0))/(6(0) + 2(0) - 2) = 2/(-2) = -1
Substitute 0 for x1, 0 for y1, and -1 for m in y - y1 = m(x - x1) to get the tangent line.
y - 0 = -1(x - 0)
y = -x
You are correct.