Partial derivatives help

Given a function z=z(x,y) defined by equation sin(xyz) = x+6y+3z.
I found dz/dx and dy/dx, its the next parts i am struggling with.

find the values of Zx(x,y) and Zy(x,y) at the point (0,1,-2) on the surface z= z(x,y) defined by above equation.

find equation of the tangent plane to the surface defined by equation sin(xyz) + -x -6y -3z at the point (0,1,-2)

Find the parameteric equation of the line perpindicular to surface in previous part at point (0,1,-2)

thank you

Let F(x,y,z) = x + 6y + 3z - sin(xyz).

So, ∂z/∂x = -(∂F/∂x) / (∂F/∂z) = -(1 - yz cos(xyz)) / (3 - xy cos(xyz))
and ∂z/∂y = -(∂F/∂y) / (∂F/∂z) = -(6 - xz cos(xyz)) / (3 - xy cos(xyz))

At (0, 1, -2):
∂z/∂x = -(1 - -2) / 3 = -1, and ∂z/∂y = -(6 - 0)/3 = -2

Hence, the equation of the tangent plane is
z - (-2) = (-1)(x - 0) + (-2)(y - 1) ==> z = -2 - (x - 0) - 2(y - 1).

Moreover, the equation of the line perpendicular to the surface at the same point is
(x - 0)/(∂F/∂x) = (y - 1)/(∂F/∂y) = (z - (-2))/(∂F/∂z) = t
==> (x - 0)/3 = (y - 1)/6 = (z + 2)/3 = t

Parametrically, this line is given by x = 3t, y = 1 + 6t, z = -2 + 3t

I hope this helps!