Let's first solve the equation:
sol[x1_, x2_, x3_] =
t3[x1, x2, x3] /.
DSolve[D[t3[x1, x2, x3], x3] +
3 D[t3[x1, x2, x3], x2] + (-1 - 9/2 x1^2) D[t3[x1, x2, x3], x1] == 0,
t3[x1, x2, x3], {x1, x2, x3}][[1]]

C[1]
here is an arbitrary function of two variables.
Let's take the second condition:
sol2[x1_, x2_, x3_] =
FullSimplify[
D[sol[x1, x2, x3], x3] + 6 D[sol[x1, x2, x3], x2] - 9 x2^2 D[sol[x1, x2, x3], x1]
]

Now, I'm not sure what is !=
supposed to mean: for the expression to not be identically equal to zero, or to never attain the value zero? If the former, then with
f[a_, b_] := a^2 + b^2
for example
sol2[1, 1, 1] /. C[1] -> f // N
3.71851
so the expression is not identically zero. Thence, the solution is
sol[x1, x2, x3] /. C[1] -> f

Here
FindInstance[sol2[x1, x2, x3] == 0 /. C[1] -> f, {x1, x2, x3}, Reals]
{{x1 -> 0, x2 -> 0, x3 -> 0}}
On the other hand, if C[1] == ArcTan
, then
sol2[1, 1, 1] /. C[1] -> ArcTan // N
-0.365403
so the solution is not identically equal to zero. Moreover,
FindInstance[sol2[x1, x2, x3] == 0 /. C[1] -> ArcTan, {x1, x2, x3}, Reals]
didn't give any instance for several minutes, so it's likely there is none and the function is non-zero everywhere.
!=
means that the expression should not be identically equal to zero, or it should for no valuesx1, x2, x3
take the value zero? $\endgroup$ – corey979 Jan 13 '17 at 22:45