Which of the following is an equation of the line tangent to the graph of \(x^2-3xy=10\) at the point \((1,-3)\) ?

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Use implicit differentiation to obtain the derivative.

$$ x^2-3xy=10 $$ $$ 2x-3\left(x\frac{dy}{dx}+y(1)\right)=0 $$

We can substitute the point \((1,-3)\) and solve for \(\frac{dy}{dx}\).

$$ 2(1)-3\left(1\left(\frac{dy}{dx}\right)+(-3)(1)\right)=0 $$ $$ 2-3\left(\frac{dy}{dx}-3\right)=0 $$ $$ \frac{dy}{dx}-3=\frac{2}{3} $$ $$ \frac{dy}{dx} = \frac{11}{3} $$

Using point slope form:

$$ y-(-3)=\frac{11}{3}(x-(1)) $$ $$ \boxed{y+3=\frac{11}{3}(x+1)} $$