Does this PDE like equation have a solution? Let $\phi:\mathbb{R}_+\times \mathbb{R}_+\to\mathbb{R}$ be a function with continuous partial derivatives in both arguments which we denote by $\partial_1$ and $\partial_2$. Consider the following differential equation:
\begin{align}
&\partial_1[\phi(t,x)+2\phi(t,y)]+\partial_2[\phi(t,x)+2\phi(t,y)]
+6\phi(t,0)-2\phi(t,x)\phi(t,y)-\phi(t,y)^2=0\;,\\[1ex]
&0\leq t<\infty\;,\;0\leq x<y\;.
\end{align}
This is not a PDE in the classical sense.
Question: Does the equation have a solution?
 A: Define
$$ F(t,x,y) = \phi_1(t,x) + 2\phi_1(t,y) + \phi_2(t,x) + 2 \phi_2(t,y) + 6 \phi(t,0) - 2\phi(t,x) \phi(t,y) - \phi(t,y)^2 $$
Your equation requires $F \equiv 0$ for $t\in [0,\infty)$ and $0\leq x < y$. For any point in the domain we must have that $\mathrm{d}F = 0$ This is equivalent to 
$$ \begin{align}
\phi_{11}(t,x) + 2\phi_{11}(t,y) + \phi_{12}(t,x) + 2 \phi_{12}(t,y) + 6 \phi_1(t,0) - 2 \phi_1(t,x) \phi(t,y) - 2 \phi(t,x) \phi_1(t,y) - 2 \phi(t,y)\phi_1(t,y) &= 0 \\
\phi_{12}(t,x) + \phi_{22}(t,x) - 2\phi_2(t,x) \phi(t,y) &= 0\\
\phi_{12}(t,y) + \phi_{22}(t,y) - \phi_2(t,y) \phi(t,x) - \phi_2(t,y) \phi(t,y) &= 0
\end{align}
$$
Consider the second of the equations. Again this holds for all points $(t,x,y)$ in the relevant domain. Taking a derivative in $y$ then requires that 
$$ \phi_2(t,x) \phi_2(t,y) = 0 $$
(note that for this we only need to assume that $\phi$ is twice continuously differentiable and not thrice!) Taking the limit $x \to y$ in the domain, by continuity we must have that $\phi_2 = 0$ everywhere!
This implies that $\phi$ is constant in its second variable, and you equation reduces to just a simple ODE
$$ \phi'(t) + 2 \phi(t) -  \phi(t)^2 = 0 $$
which we can solve to be
$$ \frac12 (\log (2 - \phi) - \log\phi) ) + C = t \implies C e^{2t} = \frac{2-\phi}{\phi} \implies \phi(t) = \frac{2}{C e^{2t} + 1}$$
where the constant $C$ on the far right hand side is determined by $\phi(0)$: 
$$ C = \frac{2}{\phi(0)} - 1$$

In summary, if we assume $\phi$ is twice continuously differentiable, then $\phi$ must be independent of its second argument and
$$ \phi(t,x) = \frac{2}{\left(\frac{2}{\phi(0,0)} - 1\right) e^{2t} + 1} $$
