Vector and parametric form of the equation of lines tangent to a surface

If we plot the graph of any $z=f(x,y)$ we will get a surface. If we take the partial derivatives at $(x_0,y_0)$ we will have two partial derivatives $f_x$ and $f_y$.

The equation of the tangent lines at that point along the $x$-axis and $y$-axis will be, respectively, $$z=z_0+f_x(x-x_0),y=y_0$$

and $$z=z_0+f_y(y-y_0),x=x_0$$

As far I know the parametric forms of the two lines will be $$(x,y,z)=(x_0+t_1,y_0,z_0+f_xt_1)$$ and $$(x,y,z)=(x_0,y_0+t_2,z_0+f_yt_2)$$

My question is, am I right? If I'm not, what is correct parametric form of the two equations? I also want to know the vector and symmetric forms of the two equations with an explanation.

The slopes of the tangent lines at $(x_0,y_0)$ pointing in the $x$ and the $y$ direction are

$$m_x=\frac{\partial }{\partial x}f_x(x,y)\mid_{x=x_0,y=y_0}=f_x(x,y)\mid_{x=x_0,y=y_0}$$ and

$$m_y=f_y(x,y)\mid_{x=x_0,y=y_0}.$$

(That is, $m_x \not =f_x(x-x_0,y_0)$ and $m_y \not =f_y(x_0,y-y_0)$.)

We have lines whose slopes are $m_x,m_y$ lie in the planes crossing the $y$ axis at $y_0$, the $x$ axis at $x_0$ and are parallel to the the $y,z$ and the $x,z$ planes, respectively. Also, these lines go through the point $(x_0,y_0,f(x_0,y_0)).$

If I am not mistaken, we are talking about the following arrangement:

If this is not the case (if we are not talking about the red and the blue lines) then I revoke my answer.

We can easily write down the equation of the line lying in the $x,z$ plane, going through the point $x_0, f(x_0,y_0)$ and whose slope is $m_x$:

$$z=xm_x-m_xx_0+f(x_0,y_0).$$

In the case of our line we have to add that $y=y_0$.

Similarly, the equation of the other line is

$$z=ym_y-m_yy_0+f(x_0,y_0), \ x=x_0.$$

In parametric form, we have

$$(x(t),y(t),z(t))=(t,y_0,m_x(t-x_0)+f(x_0,y_0))$$

and

$$(x(t),y(t),z(t))=(x_0,t,m_y(t-y_0)+f(x_0,y_0)).$$