I want to show
$$ \left( \begin{array}{ccccc} &1 &\wp(v) &\wp'(v) \\ &1 &\wp(w) &\wp'(w) \\ &1 &\wp(v+w) &-\wp'(v+w) \end{array} \right)=0 $$
where $\wp$ denotes the Weierstrass elliptic function.
I want to show
$$ \left( \begin{array}{ccccc} &1 &\wp(v) &\wp'(v) \\ &1 &\wp(w) &\wp'(w) \\ &1 &\wp(v+w) &-\wp'(v+w) \end{array} \right)=0 $$
where $\wp$ denotes the Weierstrass elliptic function.
As already mentioned in David's comment and Bruno's answer, this determinant is merely saying that the triangle formed by the three points $(\wp(v),\wp^\prime(v))$, $(\wp(w),\wp^\prime(w))$, and $(-\wp(v+w),-\wp^\prime(v+w))$ has area zero (i.e. they are collinear). (I presume the determinantal formula for the area of a triangle was taught to you in your geometry classes, no?)
In particular, you'll want to use the relation $\wp^\prime(z)^2=4\wp(z)^3-g_2 \wp(z)-g_3$. Your determinant is now equivalent to saying that a line in general position must intersect the elliptic curve $y^2=4x^3-g_2 x-g_3$ (or parametrically, $x=\wp(u)\quad y=\wp^\prime(u)$) in three points. I presume you already know how to show that a line and a cubic will have three intersection points at most. If you do the computations, you can obtain an expression for the third intersection point of a line going through the points $(\wp(v),\wp^\prime(v))$ and $(\wp(w),\wp^\prime(w))$ with the elliptic curve in terms of the coordinates of those two given points.
To get the equivalent expression in terms of $\wp(v+w)$ and $\wp^\prime(v+w)$, as Bruno mentions in his answer, one assembles the linear combination $\wp(u)+a\wp^\prime(u)+b$ and note that it has a triple pole at the origin (since $\wp^\prime(u)$ itself has a triple pole from the $-\frac2{u^3}$ term in its lattice series expansion). One can then find appropriate values of $a$ and $b$ such that $v$ and $w$ are zeroes of $\wp(u)+a\wp^\prime(u)+b$, and then find that $u=-v-w$ is a zero as well due to the properties of the Weierstrass functions within their fundamental period parallelogram. Make use of the symmetry of the two Weierstrass functions ($\wp$ is even; $\wp^\prime$ is odd), and you're golden.
This is the last of a series of exercises in Ahlfors' Complex Analysis (Pages 276-277). Those exercises are arranged in a logical sequence. I will copy and paste those that lead to this last result.
(Use (14) to show that the right-hand member is a periodic function of $z$. Find the multiplicative constant by comparing the Laurent developments.)
(Follows from (16) by taking logarithmic derivatives.)
(This is a symmetrized version of (17).)
Differentiation of (18) leads to a formula which contains $\wp''(z)$. It can be eliminated by (15) which gives $\wp" = 6\wp^2 - \frac{1}{2}g_2$. Symmetrization yields (19). Observe that this is an algebraic addition theorem, for $\wp'(z)$ and $\wp'(u)$ can be expressed algebraically through $\wp(z)$ and $\wp(u)$.)
I omit exercises 5 and 6 because they are not needed.
(14) and (15) are as follows.
$$\sigma(z+\omega_1)=-\sigma(z)e^{\eta_1(z+\omega_1/2)} \\ \sigma(z+\omega_2)=-\sigma(z)e^{\eta_2(z+\omega_2/2)} \tag{14}$$
$$\wp'(z)^2=4\wp(z)^3-g_2\wp(z)-g_3 \tag{15}$$
Below is how to derive the final result.
Differentiating (19), using $\wp''=6\wp^2-\frac{1}{2}g_2$, and symmetrising the result, we have
$$\wp’(z+u) = -\frac{1}{2} \left(\wp’(z)+\wp’(u)\right) + \frac{3}{2} \frac{\left(\wp’(z)-\wp’(u)\right)\left(\wp(z)+\wp(u)\right)}{\wp(z)-\wp(u)} - \frac{1}{4} \left(\frac{\wp’(z)-\wp’(u)}{\wp(z)-\wp(u)}\right)^3.$$
Substituting (19) and this into the determinant, we can easily show it vanishes.
(Denote the three rows of the determinant as (1), (2), (3). We replace (3) by (1)+(2)-2(3). It is then quite easy to calculate.)