Write the equation of a line tangent to the graph $3 x^{2}+2 y^{2}=10 x y+15$ at the point $(1,-1)$.
a.) $y=\frac{8}{7} x-\frac{15}{7}$
b.) $y=\frac{3}{7} x+\frac{4}{7}$
c.) $y=\frac{8}{7} x-\frac{1}{7}$
d.) $y=\frac{3}{7} x-\frac{10}{7}$

Step 1 ：First, we need to find the derivative of the given equation. The derivative of a function gives us the slope of the tangent line at any point. The equation is \(3 x^{2}+2 y^{2}=10 x y+15\).

Step 2 ：We differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\). We get \(6x + 4yy' = 10xy' + 10y\).

Step 3 ：We rearrange the equation to solve for \(y'\), the derivative of \(y\) with respect to \(x\). We get \(y' = \frac{10y - 6x}{4y - 10x}\).

Step 4 ：We substitute the point \((1,-1)\) into the derivative to find the slope of the tangent line at that point. We get \(y' = \frac{10(-1) - 6(1)}{4(-1) - 10(1)} = \frac{-16}{-14} = \frac{8}{7}\).

Step 5 ：The equation of a line is \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. We know the slope \(m\) is \(\frac{8}{7}\), and the line passes through the point \((1,-1)\), so we can substitute these values into the equation to solve for \(c\). We get \(-1 = \frac{8}{7}(1) + c\), so \(c = -\frac{15}{7}\).

Step 6 ：Thus, the equation of the line tangent to the graph at the point \((1,-1)\) is \(y = \frac{8}{7}x - \frac{15}{7}\).

Step 7 ：Checking the options, we find that this equation matches option a.) \(y=\frac{8}{7} x-\frac{15}{7}\).

Step 8 ：So, the final answer is \(\boxed{a.) y=\frac{8}{7} x-\frac{15}{7}}\).