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

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, \(\frac{d^2y}{dx^2}\). For this, we differentiate \(\frac{dy}{dx}\) with respect to \(x\).

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

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

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

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

Step 10 ：So, the second derivative, \(\frac{d^2y}{dx^2}\), 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}}