Show that a function with a tangent plane has first order partials Let $f(x,y)$ be a continuous function. Let $\phi(x,y)=a+b(x-x_0)+c(y-y_0)$ be tangent to $f$ at a point $(x_0,y_0,f(x_0,y_0))$. I want to show that $b=\frac{\partial f(x_0,y_0)}{\partial x}$, and $c=\frac{\partial f(x_0,y_0)}{\partial y}$. 
I can prove that $a=f(x_0,y_0)$ by observing that $\phi(x_0,y_0)=f(x_0,y_0)$. I know that given a continuously differentiable function $f$, the tangent plane at a point $(x_0,y_0)$ is as described above. 
What I would like to say is that the first order derivatives of $\phi$ and $f$ must be equal, but I am unsure of how to justify this, and moreover, I do not know how to justify that $f$ even has first order derivatives. A hint would be appreciated. 
Edit: By tangent, I mean that $$\lim_{(x,y)\to(x_0,y_0)}\frac{f(x,y)-\phi(x,y)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0$$
 A: Very often one defines the tangency between two graphs (or a graph and a plane) by saying that the first order derivatives must be the same. But that would give a circular reasoning for your question. 
A definition of tangency between $f(x,y)$ and $\phi(x,y)$ at $(x_0,y_0)$ without differentiability that I can think of is
$$ f(x,y) = \phi(x,y) + O(\| (x,y) - (x_0,y_0) \|^2) \text{ as $(x,y)\to(x_0,y_0)$}.$$

Edit: At the time of writing, the definition of being tangent was not given in the question, that's why I used this definition. Also, in hindsight, I could have used  $o(\|(x,y)-(x_0,y_0)\|)$.

Intuitively: the difference $f$ and $\phi$ must be (at least) of quadratic order in a neighbourhood of $(x_0,y_0)$. This definition however will imply total differentiability of $f$ at $(x_0,y_0)$, as I show you now. 
The function $f$ is totally differentiable at $(x_0,y_0)$ if there is a first order function (here the given function $\phi$) such that
$$\lim_{(x,y)\to(x_0,y_0)} \frac{\|f(x,y)-\phi(x,y)\|}{\|(x,y)-(x_0,y_0)\|} = 0. $$
By definition of the big O, there is a neighbourhood $U$ of $(x_0,y_0)$ and a constant $M>0$ such that
$$ \|f(x,y)-\phi(x,y)\| \leq M \|(x,y)-(x_0,y_0)\|^2$$
for $(x,y)$ in $U$. Hence
$$  \frac{\|f(x,y)-\phi(x,y)\|}{\|(x,y)-(x_0,y_0)\|} \leq M \|(x,y)-(x_0,y_0)\|$$
for points $(x,y) \in U$.
This implies that the abovementioned limit is zero and thus $f$ is totally differentiable in $(x_0,y_0)$. It also shows that $b=\frac{\partial f}{\partial x}(x_0,y_0)$ and $c=\frac{\partial f}{\partial y}(x_0,y_0)$.
A: Observe:
$$\frac{f(x,y_0) - f(x_0,y_0)}{x-x_0} -b = \frac{f(x,y_0) - [\phi(x_0,y_0)+b(x-x_0)]}{x-x_0}$$ $$ = \frac{f(x,y_0) - \phi(x,y_0)}{x-x_0}\to 0.$$
Therefore $\dfrac{\partial f}{\partial x}(x_0,y_0) = b.$ Similarly for $\dfrac{\partial f}{\partial y}(x_0,y_0).$
