Efficient calculation of point coordinates with elliptic curves over binary field
First, break it into a simpler problem: Divide by $u^2$ to rewrite $y^2 + uy + v = 0$ as $$(y/u)^2 + y/u + v/u^2 = 0;$$ then solve for $y/u$, and multiply the result by $u$.

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$How do you solve $t^2 + t + d = 0$ for $t$?  (Here $t = y/u$ and $d = v/u^2$.)  Suppose we are working in $\GF(2^m)$.  It is necessary that $$\Tr(d) := \sum_{i=0}^{m-1} d^{2^i} = d + d^2 + d^4 + d^8 + \dotsb + d^{2^{m - 1}} = 0,$$ where $\Tr(d)$ is the trace of $d$ over $\GF(2^m)$ as a field extension of $\GF(2)$, for if $\alpha$ is a root of $t^2 + t + d$ in $\GF(2^m)$, then $\Tr(d) = \Tr(\alpha^2 + \alpha) = \Tr(\alpha^2) + \Tr(\alpha) = \alpha^{2^m} + \alpha = 0$.  Now break this down into two cases.


$m$ is odd.  In this case, the half-trace

\begin{align*}
     H(d) := \sum_{i=0}^{\clap{(m - 1)/2}} d^{2^{2i}}
           = d + d^4 + d^{16} + \dotsb + d^{2^{m - 3}} + d^{2^{m - 1}}
   \end{align*}

is related to the trace $\Tr(d)$ by

\begin{align*}
     H(d)^2
       &amp;= \sum_{i=0}^{\clap{(m - 1)/2}} \bigl(d^{2^{2i}}\bigr)^2
        = \sum_{i=0}^{\clap{(m - 1)/2}} d^{2^{2i + 1}} \\
       &amp;= d^2 + d^8 + d^{32} + \dotsb
          + d^{2^{m - 2}} + d^{2^m} \\
       &amp;= d^2 + d^8 + d^{32} + \dotsb
          + d^{2^{m - 2}} + d \\
       &amp;= \Tr(d) + H(d) + d,
   \end{align*}

since squaring is linear in $\GF(2^m)$.  If $t^2 + t + d$ has a root so that $\Tr(d) = 0$, then we have $H(d)^2 + H(d) + d = \Tr(d) = 0$; hence $H(d)$ is a root of $t^2 + t + d$.
$m$ is even.  This case is a pain, and therefore left as an exercise for the reader who is less lazy than I am.  (Hardly anyone ever uses elliptic curves over even-power binary extension fields in cryptography; usually the power is an odd prime.)