Here it is: $M = \left\{(x,y,z)\in\mathbb{R^3}\big\vert\:\underbrace{\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}}_{f} = 1\right\}$, $(a,b,c) \geq 0$
This seems ludicrous but tempting: is the tangent space just given by $T_xM = \frac{2\,x}{a^2}+\frac{2\,y}{b^2}+\frac{2\,z}{c^2} -1 = 0$, since it's ker$(Df)$ ?
Likewise: Is it right to claim the normal space is set by: $T^\perp M = \nabla f$ ?
You have$$\nabla f(x_0,y_0,z_0)=\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},\frac{2z_0}{c^2}\right)\tag1$$and therefore the tangent plane at $(x_0,y_0,z_0)$ is the plane passing through $(x_0,y_0,z_0)$ which is orthogonal to $(1)$, which is the plane$$\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)+\frac{2z_0}{c^2}(z-z_0)=0;$$in other words, it's the plane$$\frac{x_0x}{a^2}+\frac{y_0y}{b^2}+\frac{z_0z}{c^2}=\frac{x_0^{\,2}}{a^2}+\frac{y_0^{\,2}}{b^2}+\frac{z_0^{\,2}}{c^2}.$$And the normal line is the line$$\left\{(x_0,y_0,z_0)+k\left(\frac{2x_0}{a^2},\frac{2y_0}{b^2},\frac{2z_0}{c^2}\right)\,\middle|\,k\in\Bbb R\right\}.$$