This is essentially Andrew Stacey's answer, but a bit lower level. This is the story I actually try to get my calculus 3 students to understand.
Let $F: \mathbb{R}^2 \to \mathbb{R}$. Then the derivative $D_{F,p}$ is a linear map from $D_{F,p}:\mathbb{R}^2 \to \mathbb{R}$, whose matrix with respect to the standard basis is $[ \dfrac{\partial F}{\partial x} \dfrac{\partial F}{\partial y}]$.
This is the unique linear map which satisfies $F(p+h) = F(p)+D_{F,p}(h)+Error(h)$, where $\displaystyle\lim_{h \to 0} \dfrac{|Error(h)|}{|h|} = 0$. Notice that $p$ and $h$ are both vectors in $\mathbb{R}^2$.
The cool thing about linear maps from $\mathbb{R}^n \to \mathbb{R}$ is that they look like dot products! In this case, with $h = \langle a,b \rangle$, evaluating the derivative at point $p$, then we have $D_{F,p}(\langle a,b \rangle) = [ \dfrac{\partial F}{\partial x} \dfrac{\partial F}{\partial y}] \begin{bmatrix}a \\\\ b\end{bmatrix} = \dfrac{\partial F}{\partial x}a + \dfrac{\partial F}{\partial y}b = \langle \dfrac{\partial F}{\partial x} \dfrac{\partial F}{\partial y} \rangle \cdot \langle a, b\rangle$,
(with the partials evaluated at $p$). This alternative viewpoint on the derivative is useful, because it gives a different geometric interpretation of the derivative. We call $\langle \dfrac{\partial F}{\partial x} \dfrac{\partial F}{\partial y} \rangle$ the gradient of $F$.
Now we are interested in the curve $F(x,y) = 0$. Given a point $p=(x_1,y_1)$ on this curve, the tangent direction will be the vector $h$ for which $D_{F,p}(h) = 0$, because to stay on the curve, the value of the function should not change to first order. Using the geometric interpretation in terms of dot products, we can see that $\langle \dfrac{\partial F}{\partial x} \dfrac{\partial F}{\partial y} \rangle \cdot \langle h_1, h_2\rangle = 0$, or geometrically that the gradient is perpendicular to the tangent direction!
