1. Problem: Find the derivative of the function \(x^3 + \sqrt{2x^2y} - x = 2x^2\).\n\n2. We will differentiate both sides with respect to \(x\). Note that \(y\) may be a function of \(x\), so use implicit differentiation for terms involving \(y\).\n\n3. Differentiate each term:\n- \(\frac{d}{dx}(x^3) = 3x^2\)\n- For \(\sqrt{2x^2y} = (2x^2y)^{1/2}\), use chain rule:\n \[\frac{d}{dx} (2x^2y)^{1/2} = \frac{1}{2}(2x^2y)^{-1/2} \cdot \frac{d}{dx}(2x^2y)\]\n- Differentiate inside:\n \[\frac{d}{dx}(2x^2y) = 2 \cdot (2x y + x^2 \frac{dy}{dx}) = 4xy + 2x^2 \frac{dy}{dx}\]\n- So, \(\frac{d}{dx} \sqrt{2x^2y} = \frac{1}{2\sqrt{2x^2y}} (4xy + 2x^2 \frac{dy}{dx}) = \frac{4xy + 2x^2 \frac{dy}{dx}}{2\sqrt{2x^2y}}\)\n- \(\frac{d}{dx}(-x) = -1\)\n- \(\frac{d}{dx}(2x^2) = 4x\)\n\n4. Putting it all together:\n\[3x^2 + \frac{4xy + 2x^2 \frac{dy}{dx}}{2\sqrt{2x^2y}} - 1 = 4x\]\n\n5. Multiply both sides by \(2\sqrt{2x^2y}\) to clear denominator:\n\[2\sqrt{2x^2y} (3x^2 - 1) + 4xy + 2x^2 \frac{dy}{dx} = 4x \cdot 2\sqrt{2x^2y}\]\n\n6. Rearrange to isolate \(\frac{dy}{dx}\):\n\[2x^2 \frac{dy}{dx} = 8x \sqrt{2x^2y} - 2\sqrt{2x^2y} (3x^2 - 1) - 4xy\]\n\n7. Finally, solve for \(\frac{dy}{dx}\):\n\[\frac{dy}{dx} = \frac{8x \sqrt{2x^2y} - 2\sqrt{2x^2y} (3x^2 - 1) - 4xy}{2x^2}\]\n\nThis is the implicit derivative of \(y\) with respect to \(x\).