It is a straightforward generalization of the single-variable derivative to a multivariable function. Recall that if ##f:\mathbb{R}\rightarrow\mathbb{R}##, then
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$
If we try to generalize this to two-variable functions, we have a problem with the denominator:
$$f'(x,y) = \lim_{(h,k)\to 0} \frac{f(x+h, y+k) - f(x,y)}{(h,k)}$$
The denominator is a displacement vector, and there is no consistent way to divide by a vector. However, what we want to do is divide by the length of the displacement vector, which is what we really mean when we divide by h in 1 dimension too. So let's do that instead:
$$f'(x,y) = \lim_{(h,k)\to 0} \frac{f(x+h, y+k) - f(x,y)}{||(h,k)||}$$
This looks like a derivative. But recall that in 1-dimension, the derivative was defined so that it provided the best linear approximation to f around the point at which it was taken. It was not really a number: it was a tool for finding the correct tangent line.
That is what we want in a multivariable derivative: a tool for finding the best linear approximation to f(x,y) at the point the derivative is applied. In 1 dimension, the derivative was applied to the displacement h to get the linear approximation: ##f(x + h) \approx f(x) + f'(x)\cdot h##.
A linear 2-variable approximation based on the displacement vector (h, k) would then be ##f(x, y) \approx f(x_0, y_0) + u\cdot h + v\cdot k##, where u and v are constants.
So ##f'(x,y)## somehow has to give us two numbers, u and v, to apply to the displacement vector (h,k) in order to get our linear approximation. We know the dot product gives us that expression, so f'(x,y) must be the vector (u, v). Applying f'(x,y) to a particular displacement vector (h,k) must then be done with the dot product.
But hold on. Our definition doesn't look like it is going to give us a vector:
$$f'(x,y) = \lim_{(h,k)\to 0} \frac{f(x+h, y+k) - f(x,y)}{||(h,k)||}$$
If ##f(x,y)## is a scalar-valued function, then the right side is a single number, not a 2-component vector. So whatever that is, it is not what we want.To get what we want, mathematicians made a slight adjustment to the definition. Recalling that for single variable functions, we have:
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$
we can do some slight of hand to unify the two sides. If ##f'(x)## exists, then ##\lim_{h\to 0} f'(x) = f'(x)##, so
$$\lim_{h\to 0} f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$
These two sides are just numbers, so we can do algebra:
$$0 = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} - \lim_{h\to 0} f'(x)$$
$$0 = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} - \lim_{h\to 0} \frac{f'(x)\cdot h}{h}$$
$$0 = \lim_{h\to 0} \frac{f(x+h) - f(x) - f'(x)\cdot h}{h}$$
It follows that this is an equivalent definition of the derivative, where we can see the linear approximation that it provides directly in the numerator. This motivates us to try the following definition for multivariable derivatives: ##f'(x,y)## is, at each point ##(x,y)## the unique linear function ##L(h,k)## such that:
$$0 = \lim_{(h,k)\to 0} \frac{f(x+h, y+k) - f(x,y) - L(h,k)}{||(h,k)||}$$
That is, the derivative at each point is the unique linear function that vanishes at the same rate as the function with the displacement vector. Recalling that L is a linear function of the displacement vector (h,k), it must be the case that ##L(h,k) = u\cdot h + v\cdot k##.
Use this fact in the limit, and find out what u and v must be equivalent to. You will answer your own question then. :-)