Differentiate implicitly to find $\frac{d^{2}y}{d{x}^2}$.
$$5y^2-xy+3x^2=8$$
$$\frac{d^{2}y}{d{x}^2}=$$

Step 1 ：First, we need to differentiate the given equation implicitly with respect to $x$. The given equation is $5y^2-xy+3x^2=8$.

Step 2 ：Differentiating both sides with respect to $x$, we get $10y\frac{dy}{dx}-y-x\frac{dy}{dx}+6x=0$.

Step 3 ：Rearranging the terms, we get $\frac{dy}{dx}(10y-x)=y-6x$.

Step 4 ：So, $\frac{dy}{dx}=\frac{y-6x}{10y-x}$.

Step 5 ：Now, we need to find the second derivative, $\frac{d^{2}y}{d{x}^2}$. For this, we differentiate $\frac{dy}{dx}$ with respect to $x$.

Step 6 ：Using the quotient rule, we get $\frac{d^{2}y}{d{x}^2}=\frac{(10y-x)(1-\frac{dy}{dx})-(y-6x)(10\frac{dy}{dx}-1)}{(10y-x{)}^2}$.

Step 7 ：Substituting $\frac{dy}{dx}=\frac{y-6x}{10y-x}$ into the equation, we get $\frac{d^{2}y}{d{x}^2}=\frac{(10y-x)(1-\frac{y-6x}{10y-x})-(y-6x)(10\frac{y-6x}{10y-x}-1)}{(10y-x{)}^2}$.

Step 8 ：Simplifying the equation, we get $\frac{d^{2}y}{d{x}^2}=\frac{100y^2-20xy+x^2-10y^2+20xy-x^2-10y^2+20xy-x^2+y^2-6x^2}{100y^2-20xy+x^2}$.

Step 9 ：Further simplifying, we get $\frac{d^{2}y}{d{x}^2}=\frac{80y^2-6x^2}{100y^2-20xy+x^2}$.

Step 10 ：So, the second derivative, $\frac{d^{2}y}{d{x}^2}$, of the given equation is $\frac{80y^2-6x^2}{100y^2-20xy+x^2}$.

Step 11 ：\boxed{\frac{d^2y}{dx^2} = \frac{80y^2 - 6x^2}{100y^2 - 20xy + x^2}}