Formulating the correct notion of a tangent space Let $f(x,y)$ be a polynomial in $\mathbb{R}[x,y]$.  Let $(x_1,x_2)$ be a point on the curve $Z(f(x,y)) = \{ (t_1,t_2) : f(t_1,t_2) = 0\}$. Is there an obvious reason why the tangent line at $(x_1,x_2)$ consists of the points $(x,y)$ satisfying $$\frac{\partial f}{\partial x}(x_1,y_1)(x-x_1) + \frac{\partial f}{\partial y}(x_1,y_1)(y-y_1) = 0$$ I don't know why, but I'm just not seeing this intuitively.
 A: $\newcommand{\dd}{\partial}\newcommand{\grad}{\nabla}$There are multiple ways to view things. For example:


*

*In multivariable calculus one learns that if a smooth, real-valued function $f$ has non-zero gradient $\grad f(p)$ at some point $p$, then the level set of $f$ through $p$ is a smooth hypersurface at $p$, and $\grad f(p)$ is orthogonal to the tangent space. That is, if $v = (x - x_{1}, y - y_{1})$ is a tangent vector at $p = (x_{1}, y_{1})$ to the level set, then
$$
0 = \grad f(p) \cdot v
  = \left(\frac{\dd f}{\dd x}(p), \frac{\dd f}{\dd y}(p)\right) \cdot (x - x_{1}, y - y_{1})
  = \frac{\dd f}{\dd x}(x_{1}, y_{1})(x - x_{1}) + \frac{\dd f}{\dd y}(x_{1}, y_{1})(y - y_{1}).
$$

*Differentiating the equation $c = f(x, y)$ at the point $(x_{1}, y_{1})$ gives
$$
0 = \frac{\dd f}{\dd x}(x_{1}, y_{1})\, dx + \frac{\dd f}{\dd y}(x_{1}, y_{1})\, dy.
$$
This $1$-form cuts out the subspace
$$
0 = \frac{\dd f}{\dd x}(x_{1}, y_{1})(x - x_{1}) + \frac{\dd f}{\dd y}(x_{1}, y_{1})(y - y_{1})
$$
in the tangent space to the plane at $(x_{1}, y_{1})$.

*Expanding $f$ at $(x_{1}, y_{1})$ in powers of $(x - x_{1})$ and $(y - y_{1})$ gives
\begin{align*}
f(x, y) &= f(x_{1}, y_{1}) \\
  &+ \frac{\dd f}{\dd x}(x_{1}, y_{1})(x - x_{1}) + \frac{\dd f}{\dd y}(x_{1}, y_{1})(y - y_{1}) \\
  &+ \frac{1}{2} \left[\frac{\dd^{2} f}{\dd x}(x_{1}, y_{1})(x - x_{1})^2
  + 2\frac{\dd^{2} f}{\dd x\, \dd y}(x_{1}, y_{1})(x - x_{1})(y - y_{1})
  + \frac{\dd^{2} f}{\dd y^{2}}(x_{1}, y_{1})(y - y_{1})^{2}\right]
  + \cdots.
\end{align*}
Setting this equal to $c = f(x_{1}, y_{1})$ and truncating after the first-order terms to get the tangent line gives
$$
0 = \frac{\dd f}{\dd x}(x_{1}, y_{1})(x - x_{1}) + \frac{\dd f}{\dd y}(x_{1}, y_{1})(y - y_{1}).
$$
(There are more details in my answer to Can someone provide the formal definition of the tangent line to a curve?.)
