Value of $t$ for which a curve has singular points. I want to solve this exercise.

Let $t\in\mathbb{C}$, we consider the projective curve $\mathcal{C}_{t}$ given by $X^{2}+Y^{2}+Z^{2}+t(X+Y+Z)^{2}=0$. For which $t$ does $\mathcal{C}_{t}$ have a singular point? Compute those points, their multiplicity and their tangent lines.

I know that it has to be $F(P)=0$ and $\nabla F(P)=0$ for $P$ to be a singular point. My teacher told us the answer is $t=-\frac{1}{2}$, but I don't know how to get it. I tried to do a system of equations and obtain $t$, but I don't get that value.
I think once I obtain $t$, I could find the other things.
 A: The answer I got is $t=-\frac{1}{3}$, and the value $t=-\frac{1}{2}$ also shows up at some point but has to be discarded because of one of the equations. I hope my computations are correct.
A point being singular or not depends only on a small neighborhood around the point, so we can dehomogeneize, say, setting $Z=1$. We get a nicer equation
$$ X^{2}+Y^{2}+t(X+Y+1)^{2}+1=0 $$
Now derivate with respect to $X$ for example:
$$2X+2t(X+Y+1)=0$$
This is equivalent to the equation $X+tX+tY+t=0$, and similarly we also get that $Y+tY+tX+t=0$. If we substract one equation from the other we get that necessarily $X=Y$, so we have equations
$$ X+2tX+t=0 \quad \text{and} \quad 2X^{2}+t(2X+1)^{2}+1=0$$
From the first one, $(1+2t)X=-t$. If everything above is correct, we see here that $t$ cannot be $-\frac{1}{2}$, because we get a contradiction with this equation.
Finally, since $t\neq -\frac{1}{2}$, we can divide in the previous equation to compute the value of $X$ and plug it into the other equation. The cubic terms cancel out and we get a quadratic equation on $t$ whose solutions are
$$ t=-\frac{1}{2} \quad \text{and} \quad t=-\frac{1}{3}$$
Since the first value was already discarded, the value we were looking for is the second one:
$$t=-\frac{1}{3}$$
