I am working with the following problem.
Let $X = V \left( X_1^2 - X_2X_3, X_1X_3 - X_1 \right) \subset \mathbb{C}^3$. Determine the set of all singular points of $X$.
I have proved that the irreducible decomposition of $X$ is $X = V\left( X_1, X_2 \right) \cup V\left( X_1, X_3 \right) \cup V \left( X_1^2 - X_2X_3, X_3 - 1 \right)$ and $\dim X = 1$. Here is my attempt to determine all singular points.
First we compute all intersections of different irreducible components. If $a \in V\left( X_1, X_2 \right) \cap V\left( X_1, X_3 \right)$, $a = (0,0,0)$. If $a \in V\left( X_1, X_2 \right) \cap V \left( X_1^2 - X_2X_3, X_3 - 1 \right)$, $a = (0,0,1)$. And $V\left( X_1, X_3 \right) \cap V \left( X_1^2 - X_2X_3, X_3 - 1 \right) = \varnothing$. Then we determine all smooth points in these components. For any point $a := \left( a_1, a_2, a_3 \right) \subset \mathbb{C}^3$ in $Y_1 := V \left( X_1, X_2 \right)$, the tangent space $$T_{Y_1, a} = \left\{ \left( X_1, X_2, X_3 \right) \colon X_1 = a_1, X_2 = a_2 \right\} \cong \mathbb{C},$$therefore the dimension of tangent space at any point in $Y_1$ is $1$. Hence all points in $Y_1$ are smooth. Similarly all points in $Y_2$ are smooth. For a point $a$ in $Y_3 := V \left( X_1^2 - X_2X_3, X_3 - 1 \right)$, \begin{align*} T_{Y_3,a} &= \left\{ \left( X_1, X_2, X_3 \right) \colon 2a_1X_1 - a_3X_2 - a_2 X_3 = 2a_1^2 - 2 a_2a_3, X_3 = a_3 \right\} \\ &= \left\{ \left( X_1, X_2, X_3 \right) \colon 2a_1X_1 - a_3X_2 = 2a_1^2 - a_2a_3, X_3 = a_3 \right\}. \end{align*}If $a_1 = a_3 = 0$, then $T_{Y_3, a} \cong \mathbb{C}^2$, but $a \in Y_3$ indicates $a_3 = 1$. Therefore the tangent space of any $a \in Y_3$ is the intersection of two non-degenerate plane, which is isomorphic to $\mathbb{C}^1$. So every point in $Y_3$ is of dimension $1$ and thus smooth. So $$\operatorname{Sing}(X) = \left\{ \left( 0,0,0 \right), \left( 0,0,1 \right) \right\}.$$
Is my solution correct? I appreciate every suggestion.