Can the directional derivative fail to be linear? Is it possible for the directional derivative for a function $f$ in the direction of a vector $v$, $D_vf(x) = \lim_{h \to 0} \frac{f(x + hv) - f(x)}h$ to exist for every vector $v$, and yet $v \mapsto D_vf(x)$ fails to be linear? Here we have a function which seems to be displaying that very phenomenon:
$$f(x,y) = \frac{32x^3}{x^2 + y^2} - \frac{16x^5}{(x^2 + y^2)^2} - 14x$$

As you can see, if you follow the map at the origin in the direction of any vector, it appears to follow a straight line, i.e. the derivative in that direction is constant in the direction of $v$. Yet the "slope" in each direction is clearly not behaving in a linear manner. Is this really the case? Or is it just a misleading graph? I'm inclined to believe the former, but what I'm looking for is a rigorous argument.
Further, under what conditions might a derivative fail to be linear? Or, what are necessary conditions to ensure that a proof that "the directional derivative is linear" actually works?
 A: The map $ v\mapsto D_vF(x_0)$ is linear when the function $f$ is differentiable at $x_0$. In that case, there exists a linear map $ df(x_0)$ such that $D_vF(x_0)=\langle df(x_0),v\rangle $. There exist functions which have all the directional derivatives $D_vF(x_0)$ at a point $ x_0$, but fail to be differentiable, in particular the previous map is non linear.
A: Certainly this can happen. The classic example is $f(x,y) = xy/(x^2+y^2).$ If we let $v=(a,b),$ then $D_v(0,0) = ab/(a^2+b^2).$ The last expression is not a linear function of $(a,b).$
A: Let's look at your function
$$f(x,y) = \frac{32x^3}{x^2+y^2} - \frac{16x^5}{(x^2+y^2)} - 14 x$$
I'll use the notation $D_x = D_{(1,0)}$ and $D_y = D_{(0,1)}$.  Then,
$$D_x f(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \frac{32h^3}{h^3} - \frac{16h^5}{h^5} - 14\frac{h}{h} = 2$$
and
$$D_y f(0,0) = \lim_{h\to0}\frac{f(0,h) - f(0,0)}{h} = 0$$
But,
$$D_{(1,1)}f(0,0) = \lim_{h\to0}\frac{f(h,h)-f(0,0)}{h}= \frac{32h^3}{2h^3}-\frac{16h^5}{4h^5} - 14\frac{h}{h} = -2\not= D_xf(0,0) + D_yf(0,0)$$
So $D_v$ is not linear in $v$ at $(0,0)$.

In order for the directional derivative to be linear in $v$ at a point $x$, it is sufficient that the partial derivative $D_x$ and $D_y$ (and $D_z$, etc. for higher dimensions) be continuous at $x$.  To see why, observe that
$$D_{(v_x, v_y)} f(x) = \lim_{h\to0} \frac{f(x+hv_xe_1+hv_ye_2) - f(x)}{h} = \lim_{h\to 0} \frac{f(x+hv_xe_1+hv_2e_2) - f(x+hv_ye_2)}{h} + \lim_{h\to0} \frac{f(x+hv_ye_2) - f(x)}{h}$$
The second term we recognize as $D_{(0,v_y)}f(x)= v_yD_yf(x)$, while by the Mean Value Theorem we can re-write the first term as
$$\lim_{h\to 0} v_xD_xf(x+hv_ye_2+ce_1)$$
for some $c$ between $0$ and $h$.  But, by the continuity of $D_xf$ near $x$, this limit is simply $v_xD_xf(x)$.  Thus,
$$D_{(v_x,v_y)}f(x) = v_xD_xf(x) + v_yD_xf(x)$$
is linear in $v$.
