Given the condition $$ \frac{\partial^2 X}{\partial Y^2} (\delta Y)^2 + 2 \frac{\partial^2 X}{\partial Y \partial Z} \delta Y \delta Z + \frac{\partial^2 X}{\partial Z^2} (\delta Z)^2 > 0.$$
If $\delta Y$ and $\delta Z$ are arbitrary, why must the following two conditions hold:
$$ \frac{\partial^2 X}{\partial Y^2} > 0,$$
$$ \frac{\partial^2 X}{\partial Y^2} \frac{\partial^2 X}{\partial Z^2} - \left(\frac{\partial^2 X}{\partial Y \partial Z} \right)^2 > 0 \, \,?$$
First, the expression is zero for $\delta Y = \delta Z = 0$, to you can only ask for conditions for it to be positive for arbitrary $(\delta Y, \delta Z) \ne (0, 0)$. Changing the notation, the question is: Under what conditions is $$ f(x, y) = A x^2 + 2B x y + C y^2 > 0 $$ for arbitrary $(x, y) \ne (0, 0)$?
$A = f(1, 0)$ must be positive. Completing the square gives $$ f(x, y) = A \left( (x+\frac BA y)^2 + \frac{AC-B^2}{A^2} y^2 \right) $$ which shows that necessary and sufficient conditions are $A > 0$ and $AC - B^2 > 0$.
Alternatively, $$ f(x, y) = (x, y) \begin{pmatrix} A & B \\ B & C \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$ is positive for all nonzero vectors $(x, y)$ if and only if the matrix $\begin{pmatrix} A & B \\ B & C \end{pmatrix}$ is positive definite, and a necessary and sufficient condition is that all its leading principal minors are all positive.
Suppose $\delta Y\neq0$. Let $k=\frac{\delta Z}{\delta Y}$. Then the condition is equivalent to $$ \frac{\partial^2 X}{\partial Y^2} + 2 \frac{\partial^2 X}{\partial Y \partial Z} k + \frac{\partial^2 X}{\partial Z^2}k^2 > 0 \tag{1}$$ for any $k$. Clearly the LHS of (1) is a quadratic function of $k$. Since it is always positive, one must have $\frac{\partial^2 X}{\partial Y^2}>0$ and its discriminant $\Delta$ to be negative. The latter implies $$\frac{\partial^2 X}{\partial Y^2} \frac{\partial^2 X}{\partial Z^2} - \left(\frac{\partial^2 X}{\partial Y \partial Z} \right)^2<0$$